.ENGINEERING
THERMODYNAMICS
BY
CHARLES EDWARD LUCKE, PH.D., 
f> $
Professor of Mechanical Engineering in Columbia University, New York City
McGEAWHILL BOOK COMPANY
239 WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
1912
PREFACE
CALCULATIONS about heat as a form of energy, and about work, another
related form, both of them in connection with changes in the condition of all
sorts of substances that may give or take heat, and perform or receive work
while changing condition, constitute the subject matter of this book. The
treatment of the subject matter of this text is the result of personal experience
in professional engineering practice and teaching students of engineering at
Columbia University.
Even a brief examination of the conditions surrounding changes in sub
stances as they gain or lose heat, do work or have work done on them, and of
the corresponding relations between heat and work as forms of energy independ
ent of substances, will convince any one that the subject is one of great com
plexity. Accordingly the simplicity needed for practical use in the industries
can be reached only by a consideration of a great mass of subtopics and data.
That the doing of work, and the changes in heat content of substances were
related phenomena, and that these relations when formulated, would con
stitute a branch of science, was conceived about a half century ago, and the
science was named Thermodynamics. The Engineer Rankine, who helped
to create it, defined thermodynamics as " the reduction of the laws according
to which such phenomena took place to a physical theory or corrected system
of principles." Since Rankine's time thermodynamics has become a very
highly developed science and has proved of great assistance in the formu
 lation of modern physical chemistry, and to those branches of engineering
that are concerned with heat. Unfortunately, as thermodynamics developed
[ as a separate subject it did not render proportionate service to engineering,
i which itself developed even more rapidly in the same period under the guidance
of men whose duty it was to create industrial apparatus and make it work
properly, and who had little or no time to keep in touch with purely scientific
: advances or to interpret such advances for utilitarian ends. Thermodynamics
proper is concerned with no numerical quantities nor with any particular
substance nor for that matter with any actual substances whatever, but it is a
physical theory of energy in relation to matter as a branch of natural philosophy.
Engineering, however, is concerned with real substances, such as coal, steam,
arid gases and with numerical quantities, horsepowers of engines, 'temper
atures of steam, the heats of combustion of oils, so that alone, the principles
of thermodynamic philosophy will not yield a solution of a practical problem,
v
271310
vi PREFACE
be it one of design or one of analysis of test performance of actual heat machine
or thermal apparatus. It is the province of engineering thermodynamics to
guide numerical computation on thermal problems for real substances being
treated in real apparatus. Its field, while including some of that of pure
thermodynamics, extends far beyond the established provinces of that subject
and extends to the interpretation of all pertinent principles and facts for purely
useful purposes. Engineering thermodynamics, while using whatever prin
ciples of pure thermodynamics may help to solve its problems, must rely on
a great mass of facts or relations that may not yet have risen to the dignity
of thermodynamic laws. The workers in shops, factories, power plants or
laboratories engaged in designing or operating to the best advantage machines
and apparatus using heat with all sorts of substances, have developed great
quantities of rules, methods and data that directly contribute to the ends sought.
While for each class or type of apparatus there has grown up a separate set
of data and methods in which much is common to several or all groups, not
nearly so much assistance is rendered by one to another as should be by a proper
use of engineering thermodynamics, which applies methods, principles and
conclusion to all related problems. Classes of apparatus about which such
groups of methods of analysis or synthesis, or collections of special data
have developed and which it is province of engineering thermodynamics to
unify so far as may be, include air compressors, and compressed air engines,
reciprocating steam engines, steam turbines, steam boilers, coal, oil and gas
fired furnaces, gasifiers of coal and oil, gas producers, gas engines, complete
steam or gas power plants, mechanical refrigeration and icemaking plants and
chemical factory equipment, or more generally, machinery and apparatus for
heating and cooling, evaporating and condensing, melting and freezing, moisten
ing and drying, gasification and combustiom.
The nature of the subject and its division are better indicated by the
classes of problems to be solved by its aid or the contributions expected of it
than by the kinds of apparatus to which they apply. Probably its broadest
contribution is the establishment of limits of possible performance of heat
apparatus and machines. These limits will show what might be expected of
a steam engine, gas engine or refrigerating machine when its mechanism is quite
perfect and thus they become standards of reference with which actual per
formance can be compared, and a measure of the improvements yet possible.
These same methods and practices are applicable to the analysis of the operat
ing performance of separate units and complete plants to discover the amount
of energy being lost, how the total amount is divided between the different
elements of the apparatus, which of the losses can be prevented and how, and
finally which are absolutely unavoidable. This sort of analysis of the per
formance of thermal apparatus is the first step to be taken by the designer
or manufacturer to improve the machine that he is creating for sale, and is
essential to the purchaser and user of the machine, who cannot possibly keep
it in the best operating condition without continually analyzing its performance
and comparing results with thermodynamic possibilities.
PREFACE vii
The subject naturally divides into three general parts, the first dealing
with the conditions surrounding the doing of work without any consideration
of heat changes, the second heat gains and losses by substances without reference
to work involved and the third, transformation of heat into work or work into
heat in conjunction with changes in the condition of substances. The first
part applies to the behavior of fluids in the cylinders of compressors and engines.
The second part is concerned with the development of heat by combustion,
its transmission from place to place, and the effect on the physical condition of
solids, liquids, gases with their mixtures, solutions and reactions. The third
part is fundamental to the efficient production of power by gases in internal
combustion gas engines or compressedair engines and by steam or other vapors
in steam engines and turbines, and likewise as well to the production of
mechanical refrigeration by ammonia, carbon dioxide and other vapors.
Accordingly, the six chapters of the book treat these three parts in order.
The first three chapters deal with work without any particular reference to
heat, the second two with heat, without any particular reference to work,
while the last is concerned with the relation between heat and work. After
establishing in the first chapter the necessary units and basic principles for
fixing quantities of work, the second chapter proceeds at once to the determina
tion of the work done in compressor cylinders and the third, the available work
in engine cylinders, in terms of all the different variables that may determine
the work for given dimensions of cylinder or for given quantities of fluid. There
is established in these first three chapters a series of formulas directly applicable
to a great variety of circumstances met with in ordinary practice. All are
derived from a few simple principles and left in such form as to be readily
available for numerical substitution. This permits of the solution of numerical
problems on engine and compressor horsepower, fluid consumption or capacity
with very little labor or time, although it has required the expansion of the
subject over a considerable number of pages of book matter. A similar pro
cedure is followed in the succeeding chapters, formulas and data are developed
and placed always with a view to the maximum clearness and utility. The
essential unity of the entire subject has been preserved in that all the important
related subjects are treated in the same consistent manner and at sufficient
length to make them clear. When no general principles were available for
a particular solution there has been no hesitation in reverting to specific data.
The subject could have been treated in a very much smaller space with less
labor in book writing but necessitating far greater labor in numerical work
on the part of the user. This same aim, that is, the saving of the user's
time and facilitating the arrival at numerical answers, is responsible for the
insertion of a very considerable number of large tables, numerous original
diagrams and charts, all calculated for . the purpose and drawn to scale.
These, however, take a great deal of room but are so extremely useful in
everyday work as to justify any amount of space thus taken up. For
the sake of clearness all the steps in the derivation of any formula used
arc given, and numerical examples are added to illustrate their meaning
viii PREFACE
and application, This also requires a considerable amount of space but
out it the limitations of the formulas would never be clear nor could a student
learn the subject without material assistance. Similarly, space has been used
in many parts of the book by writing formulas out in words instead of express
ing them in symbols. This saves a great deal of time and labor in hunting up
the meaning of symbols by one who desires to use an unfamiliar formula involv
ing complex quantities, the meaning of which is often not clear when it is entirely
symbolic. Thus, in the discussion of steam boiler capacity and efficiency, a
dozen or more pages are taken up with formulas that could have been con
centrated in a single page were symbols used entirely, but only at the sacrifice
of clearness and utility. Where in the derivation of a new formula or in the
treatment of a new subject, reference to an old formula or statement is needed
and important, repetition is resorted to, rather than mere reference, so that
the new topic may be clear where presented, without constantly turning the
pages of the book. It will be found, therefore, that while the size of the book
is unusually large it will be less difficult to study than if it were short.
As a text the book may be used for courses of practically any length, but
it is not int ended. Vhat in any course on the subject every page of the book shall
be used as assipued text. In the new graduate course in mechanical engineering
at Columbia University, about threefourths of the subject matter of the book
will be so used for a course of about one hundred and twenty periods of one
hour each. All of the book matter not specifically assigned as text or reference
in a course on engineering thermodynamics in any school may profitably be
taken up in courses on other subjects, serving more or less as a basis for them.
It is therefore adapted to courses on gas power, compressed air, steam turbines,
steam power plants, steam engine design, mechanical refrigeration, heating
and ventilation, chemical factory equipment, laboratory practice and research.
Whenever a short course devoted to engineering thermodynamics alone is
desired, the earlier sections of each chapter combined in some cases with the
closing sections, may be assigned as text. In this manner a course of about
thirty hours may be profitably pursued. This is a far better procedure than
using a short text to fit a short course, as the student gets a better perspective,
and may later return to omitted topics without difficulty.
The preparation of the manuscript involves such a great amount of labor,
that it would never have been undertaken without the assurance of assistance
by Mr. E. D. Thurston, Jr., a fellow instructor at Columbia, in checking text
and tables, calculating diagrams, writing problems and working examples.
This help has been invaluable and is gratefully acknowledged. Recognition
is also due for material aid rendered by Mr. T. M. Gunn in checking and in
some cases deriving formulas, more especially those of the first three chapters.
In spite, however, of all care to avoid errors it is too much to expect complete
success in a new work of this character, but it is hoped that readers finding
errors will point them out that future editions may be corrected.
C. E. L.
COLUMBIA UNIVERSITY, NEW YORK, September, 1912,
CONTENTS
CHAPTER I. WORK AND POWER. GENERAL PRINCIPLES
PAGE
1. Work defined 1
2. Power denned 2
3. Work in terms of pressure and volume 3
4. Work of acceleration and resultant velocity 6
5. Graphical representation of work 8
6. Work by pressure volume change 10
7. Work of expansion and compression . 13
8. Values of exponent s defining special cases of expansion or compression 20
9. Work phases and cycles, positive, negative and net work 24
10. Work determination by mean effective pressure .' 31
11. Relation of pressurevolume diagrams to indicator cards 34
12. To find the clearance 37
.3. Measurement of areas of PV diagrams and indicator cards 43
.4. Indicated horsepower 44
.5. Effective horsepower, brake horsepower, friction horsepower, mechanical efficiency,
efficiency of transmission, thermal efficiency : 47
.6. Specific displacement, quantity of fluid per hour, or per minute per I.H.P 49
.7. Velocity due to free expansion by PV method 52
8. Weight of flow through nozzles by PV method 55
9. Horsepower of nozzles and jets, by PV method 57
CHAPTER II. WORK OF COMPRESSORS. HORSEPOWER and CAPACITY OF AIR, GAS AND
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS
1. General description of structures and processes 73
2. Standard reference diagrams or PV cycles for compressors and methods of analysis
of compressor work and capacity 75
3. Singlestage compressor, no clearance, isothermal compression, Cycle I. Work,
capacity, and work per cubic foot in terms of pressures and volumes 81
4. Singlestage compressor with clearance, isothermal compression, Cycle II. Work,
capacity, and work per cubic foot in terms of pressures and volumes 85
5. Singlestage compressor, isothermal compression. Capacity, volumetric efficiency,
work, mean effective pressure, hor&epower and horsepower per cubic foot of
substance, in terms of dimensions and cylinder clearance 87
6. Singlestage compressor, no clearance, exponential compression, Cycle III. Work,
capacity and work per cubic foot, in terms of pressures and volumes 91
7. Singlestage compressor with clearance, exponential compression, Cycle IV. Work,
capacity, and work per cubic foot in terms of pressures and volumes 96
8. Singlestage compressor, exponential compression. Relation between capacity,
volumetric efficiency, work, mean effective pressure, horsepower and horse
power per cubic foot of substance and the dimensions of cylinder and clearance . . 98
ix
x CONTENTS
PAGE
9. Twostage compressor, no clearance, perfect intercooling, exponential compression,
bestreceiver pressure, equality of stages. Cvcle V. Work and capacity in terms
of pressures and volumes 103
10. Twostage compressor, with clearance, perfect intercooling 6x7 < nitial compression,
bestreceiver pressure, equality of stages, Cycle VI. Work and capacity in terms
of pressures and volumes 109
11. Twostage compressor, any receiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure and horsepower, in terms of
dimensions of cylinders and clearances 113
12. Twostage compressor, with bestreceiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure and horsepower in terms
of dimensions of cylinders and clearances 120
13. Threestage compressor, no clearance, perfect intercooling exponential compres
sion, best two receiver pressures, equality of stages, Cycle VII. Work and
capacity, in terms of pressures and volumes 125
14. Threestage compressor with clearance, perfect intercooling exponential compression,
bestreceiver pressures, equality of stages, Cycle VIII. Work and capacity in
terms of pressures and volumes 131
15. Threestage compressor, any receiver pressure, exponential compression. Capacity,
volumetric efficiency, work, mean effective pressure, and horsepower in terms
of dimensions of cylinders and clearances 135
16. Threestage compressor urith bestreceiver pressures, exponential compression. Capac
ity, volumetric efficiency, work, mean effective pressure and horsepower in terms
of dimensions of cylinders and clearances 143
17. Comparative economy or efficiency of compressors 148
18. Conditions of maximum work of compressors 151
19. Compressor characteristics 153
20. Work at partial capacity in compressors of variable capacity 160
21. Graphic solution of compressor problems 168
CHAPTER III. WORK OF PISTON ENGINES. HORSEPOWER AND CONSUMPTION OF
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER GAS OR VAPOR
UNDER PRESSURE
1. Action of fluid in single cylinders. General description of structure and processes. . 187
2. Simple engines. Standard reference cycles or PV diagrams for the work of expansive
fluids in a single cylinder 192
3. Work of expansive fluid in single cylinder without clearance. Logarithmic expan
sion, Cycle I. Mean effective pressure, horsepower and consumption of simple
engines 197
4. Work of expansive fluid in single cylinder without clearance. Exponential expan
sion, Cycle II. Mean effective pressure, horsepower and consumption of simple
engines 205
5. Work of expansive fluid in single cylinder with clearance. Logarithmic expansion
and compression; Cycle III. Mean effective pressure, horsepower, and con
sumption of simple engines 208
6. Work of expansive fluid in single cylinder with clearance; exponential expansion
and compression, Cycle IV. Mean effective pressure, horsepower and consumption
of simple engines 219
7. Action of fluid in multipleexpansion cylinders. General description of structure
and processes 225
8. Standard reference cycles or PV diagrams for the work of expansive fluids in two
cylinder compound engines 235
CONTENTS xi
PAGE
9. Compound engine with infinite receiver. Logarithmic law. No clearance, Cycle
V. General relations between pressures, dimensions, and work 256
0. Compound engine with infinite receiver. Exponential law. No clearance, Cycle
VI. General relations between pressures, dimensions and work 268
1. Compound engine with finite receiver. Logarithmic law. No clearance, Cycle.
VII. General relations between dimensions and work when H.P. exhaust and
L.P. admission are not coincident. . . 274
2. Compound engine with finite receiver. Exponential law, no clearance, Cycle VIII.
General relations between pressures, dimensions, and work, when high pressure
Exhaust and lowpressure admission are independent 287
3. Compound engine without receiver. Logarithmic law. No clearance, Cycle IX.
General relations between dimensions and work when highpressure exhaust
and lowpressure admission are coincident 292
4. Compound engine without receiver. Exponential law. No clearance, Cycle X.
General relations between dimensions and work when highpressure exhaust
and lowpressure admission are coincident 301
5. Compound engine with infinite receiver. Logarithmic law, with clearance and
compression, Cycle XI. General relations between pressures, dimensions and
work 306
6. Compound engine with infinite receiver. Exponential law, with clearance and
compression, Cycle XII. General relations between pressures, dimensions and
work 319
7. Compound engine with finite receiver. Logarithmic law, with clearance and com
pression, Cycle XIII. General relations between pressures, dimensions, and
work when H.P. exhaust and L.P. admission are independent 325
8. Compound engine with finite receiver. Exponential law, with clearance and com
pression, Cycle XIV. General relations between pressures, dimensions, and
work when H.P. exhaust and L.P. admission are independent 335
9. Compound engine without receiver. Logarithmic law, with clearance and com
pression, Cycle XV. General relations between pressures, dimensions, and
work when H.P. exhaust and L.P. admission are coincident 339
0. Compound engine without receiver. Exponential law, with clearance and compres
sion, Cycle XVI. General relations between pressures, dimension, and work,
when H.P. exhaust and L.P. admission are coincident , 346
1 1. Tripleexpansion engine with infinite receiver. Logarithmic law. No clearance,
Cycle XVII. General relations between pressures, dimensions and work 349
2. Multipleexpansion engine. General case. Any relation between cylinders and
receiver. Determination of pressurevolume diagram and work, by graphic
methods  357
3. Mean effective pressure, engine horsepower, and work distribution and their vari
ation with valve movement and initial pressure. Diagram distortion and diagram
factors. Mechanical efficiency 363
1. Consumption of steam engines and its variation with valve movement and initial
pressure. Best cutoff as affected by condensation and leakage 371
5. Variation of steam consumption with engine load. The Willans line. Most eco
nomical load for more than one engine and best load division 381
3. Graphical solution of problems on engine horsepower and cylinder sizes 387
xii CONTENTS
CHAPTER IV. HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS
BETWEEN HEAT CONTENT OF SUBSTANCES AND PHYSICALCHEMICAL STATE
PAGE
1. Substances and heat effects important in engineering. . ( 398
2. Classification of heating processes. Heat addition and abstraction with or without
temperature change, qualitative relations 401
3. Thermometry based on temperature change, heat effects. Thermometer and abso
lute temperature scales 407
4. Calorimetry based on proportionality of heat effects to heat quantity. Units of
heat and mechanical equivalent 415
5. Temperature change relation to amount of heat for solids, liquids, gases and vapors
not changing state. Specific heats 419
6. Volume or density variation with temperature of solids, liquids, gases and vapors
not changing state. Coefficients of expansion. Coefficients of pressure change
for gases and vapors 435
7. Pressure, volume, and temperature relations for gases. Perfect and real gases 438
8. Gas density and specific volume and its relation to molecular weight and gas constant . 446
9. Pressure and temperature relations for vapor of liquids or solids. Vaporization,
sublimation and fusion curves. Boiling and freezingpoints for pure liquids
and dilute solutions. Saturated and superheated vapors 451
10. Change of state with amount of heat at constant temperature. Latent heats of
fusion and vaporization. Total heats of vapors. Relation of specific volume
of liquid and of vapor to the latent heat 467
11. Gas and vapor mixtures. Partial and total gas and vapor pressures. Volume,
weight and gas constant relations. Saturated mixtures. Humidity 481
12. Absorption of gases by liquids and by solids. Relative volumes and weights with
pressure and temperature. Heats of absorption and of dilution. Properties
of aqua ammonia 493
13. Combustion and related reactions. Relative weights and volumes of substances
and elements before and after reaction 506
14. Heats of reaction. Calorific power of combustible elements and of simple chemical
compounds. B.T.U. per pound and per cubic foot 516
15. Heat transmission processes. Factors of internal conduction, surface resistance,
radiation and convection 528
16. Heat transmission between separated fluids. Mean temperature differences, coeffi
cients of transmission 538
17. Variation in coefficient of heat transmission due to kind of substance, character of
separating wall and conditions of flow v 555
CHAPTER V. HEATING BY COMBUSTION. FUELS, FURNACES, GAS PRODUCERS AND
STEAM BOILERS
1. Origin of heat and transformation to useful form. Complexity of fuels as sources
of heat. General classification, solid, liquid, gaseous, natural and artificial 644
2. Natural solid fuels, wood peat, lignite, bituminous and anthracite coals. Chemical
and physical properties. Classifications based on ultimate and proximate analysis
and on behavior on heating 649
3. Calorific power of coals and the combustible of coals. Calculation of calorific power
from ultimate and proximate analyses. Calorific power of the volatile 662
4. Mineral oil and natural gas fuel. Chemical and physical properties. Calorific
power direct and as calculated for oils from ultimate analysis or from density,
and for gas from sum of constituent gases 67C
CONTENTS xiii
PAGE
5. Charcoal, coke, coke oven and retort coal gas as products of heating wood and coal.
Chemical, physical, and calorific properties per pound. Calorific power of gases
per cubic foot in terms of constituent gases. Yield of gas and coke per pound
coal 675
6. Distillate oils, kerosene and gasoline, residue oils, and oil gas as products of heating
mineral oils. Chemical, physical, and calorific properties. Calorific power of frac
tionated oils in terms of, (a) carbon and hydrogen; (6) density per pound, and
estimated value per cubic foot vapor. Calorific power of oil gas per pound and per
cubic foot in terms of constituent gases. Yield of distillates and oil gas 685
7. Gasification of fixed carbon and coke by airblast reactions, producing ah* gas, and
blastfurnace gas. Comparative yield per pound coke and air. Sensible heat
and heat of combustion of gas. Relation of constituents in gas. Efficiency
of gasification 695
8. Gasification of fixed carbon, coke, and coal previously heated, by steamblast reac
tions, producing water gas. Composition and relation of constituents of water
gas, yield per pound of steam and coal. Heat of combustion of gas and limitation
of yield by negative heat of reaction 710
9. Gasification of coals by steam and air blasts, resulting in producer gas. Composition
and relation of constituents of producer gas, yield per pound of fixed carbon, air
and steam. Modification of composition by addition of volatile of coal. Heat
of combustion of gas, sensible heat, and efficiency of gasification. Horsepower
of gas producers 719
10. Combustion effects. Final temperature, volume and pressure for explosive and
nonexplosive combustion. Estimation of air weights and heat suppression
due to CO in products from volumetric analysis 740
11. Temperature of ignition and its variation with conditions. Limits of proportion
air gas neutral, or detonating gas and neutral, for explosive combustion of mix
tures. Limits of adiabatic compression for selfignition of mixtures 758
12. Rate of combustion of solid fuels with draft. Propagation rates, uniform and
detonating for explosive gaseous mixtures 765
13. Steamboiler evaporative capacity and horsepower. Horsepower units, equivalent
rates of evaporation and of heat absorption. Factors of evaporation. Relation
between absorption rates and rates of heat generation. Influence of heating and
grate surface, calorific power of fuels, rates of combustion and furnace losses. . . . 773
14. Steamboiler efficiency, furnace and heatingsurface efficiency. Heat balances and
variation in heat distribution. Evaporation and losses per jpound of fuel 796
CHAPTER VI. HEAT AND WORK. GENERAL RELATIONS BETWEEN HEAT AND WORK.
THERMAL EFFICIENCY OF STEAM, GAS, AND COMPRESSEDAIR ENGINES. FLOW
OF EXPANSIVE FLUIDS. PERFORMANCE OF MECHANICAL REFRIGERATING SYSTEMS
1. General heat and work relations. Thermal cycles. Work and efficiency deter
mination by heat differences and ratios. Graphic method of temperature
entropy heat diagram , 874
2. General energy equation between heat change, intrinsic energy change, and work
done. Derived relations between physical constants for gases and for changes
of state, solid to liquid, and liquid to vapor 882
3. Quantitative relations for primary thermal phases, algebraic, and graphic to PV,
and T3> coordinates. Constancy of PV, and T for gases and vapors, wet, dry
and superheated 892
4. Quantitative relations for secondary thermal phases. Adiabatics for gases and
vapors. Constant quality, constant total heat, and logarithmic expansion lines
for steam. . 904
xiv CONTENTS
5. Thermal cycles representative of heatengine processes. Cyclic efficiency. A
reference standard for engines and fuelburning power systems. Classification
of steam cycles
6. The Rankine cycle. Work, mean effective pressure, jet velocity, water rate, heat
consumption and efficiency of steam Cycle I. Adiabatic expansion, constant
pressure, heat addition and abstraction, no compression 936
7. Carnot cycle and derivatives. Work, mean effective pressure, water rate, heat
consumption and efficiency of steam Cycle II. Adiabatic expansion and com
pression, constant pressure heat addition and abstraction 957
8. Gas cycles representative of ideal processes and standards of reference for gas
engines 970
9. Brown, Lenoir, Otto and Langen noncompression gascycles. Work, mean
effective pressure, volume and pressure ranges, efficiency, heat and gas con
sumption 978
10. Stirling and Ericsson gas cycles. Work, efficiency and derived quantities for
isothermal compression with and without regenerators 993
11. Otto, Atkinson, Brayton, Diesel, and Carnot gas cycles. Work, efficiency and
derived quantities for adiabatic compression gas cycles 1006
12. Comparison of steam and gas cycles with the Rankine as standard for steam, and
with the Otto and Diesel, as standard for gas. Relations of Otto and Diesel
to Rankine cycle. Conditions for equal efficiency 1031
13. Gas cycle performance as affected by variability of the specific heats of gases,
applied to the Otto cycle 1035
14. Actual performance of Otto and Diesel gas engines, and its relation to the cyclic.
Diagram factors for mean effective pressure and thermal efficiency. Effect
of load on efficiency. Heat balance of gas engines alone and with gas producers 1042
15. Actual performance of piston steam engines and steam turbines at their best load
and its relation to the cyclic. Effect on efficiency of initial pressure, vacuum,
superheat, jacketing, and reheating. Heat balances of steam power plants .... 1062
16. Flow of not water, steam and gases through orifices and nozzles. Velocity, weight
per second, kinetic energy, and force of reaction of jets. Nozzle friction and
reheating and coefficient of efflux. Relative proportions of series nozzles for
turbines for proper division of work of expansion 1083
17. Flow of expansive fluids under small pressure drops through orifices, valves, and
Venturi tubes. Relation between loss of pressure and flow. Velocity heads
and quantity of flqw by Pitot tubes 1097
18. Flow of gases and vapors in pipes, flues, ducts, and chimneys. Relation between
quantity of flow and loss of pressure. Friction resistances. Draught and
capacity of chimneys 1111
19. Thermal efficiency of compressedair engines alone and in combination with air
compressors. Effect of preheating and reheating. Compressor suction heating,
and volumetric efficiency. Wall action 1127
20. Mechanical refrigeration, general description of processes and structures. Thermal
cycles and refrigerating fluids. Limiting temperatures and pressures 1142
21. Performance of mechanical refrigerating cycles and systems. Quantity of fluid
circulated per minute per ton refrigeration, horsepower, and heat supplied
per ton. Refrigeration per unit of work done and its relation to thermal effi
ciency of the system 1157
LIST OF TABLES
NO. PAGE
1. Conversion table of units of distance 62
2. Conversion table of units of surface 62
3. Conversion table of units of volume 62
4. Conversion table of units of weights and force 63
5. Conversion table of units of pressure 63
6. Conversion table of units of work 64
7. Conversion table of units of power 64
8. Units of velocity 64
9. Barometric heights, altitudes and pressures 65
10. Values of s in the equation PV S = constant for various substances and conditions . . 67
11. Horsepower per pound mean effective pressure 68
12. Ratio of cutoffs in the two cylinders of the compound engine to give equal work
for any receiver volume 284
13. Piston positions for any crank angle 395
14. Values for x for use in Heck's formula for missing water 396
15. Some actual steam engine dimensions 396
16. Fixed temperatures 411
17. Fahrenheit temperatures by hydrogen and mercury thermometers 414
18. Freezingpoint of calcium chloride brine ; 425
19. Specific heat of sodium chloride brine 427
20. Specific heat and gas constants, 431
21. The critical point 453
22. Juhlin's data on the vapor pressure of ice 456
23. Tamman's value on fusion pressure and temperature of waterice 456
24. Lowering of freezingpoints 465
25. Berthelot's data on heat for complete dilution of ammonia solutions 500
26. Air required for combustion of various substances 515
27. Radiation coefficients 535
28. Coefficients of heat transfer 550
29. Temperatures, Centigrade and Fahrenheit 571
30. Heat and power conversion table 573
31. Specific heat of solids 574
32. Specific heats of liquids 576
33. Baumespecific gravity scale 577
34. Specific heats of gases 578
35. Coefficient of linear expansion of solids 580
36. Coefficient of cubical expansion of solids 581
37. Coefficient of volumetric expansion of gases and vapors at constant pressure 582
xv
xv i LIST OF TABLES
NO, PAGE
38. Coefficient of pressure rise of gases and vapors at constant volume 583
39. Compressibility of gases by their isothermals 584
40. Values of the gas constant R 584
41. Density of gases 585
42. International atomic weights 586
43. Melting or freezingpoints 586
44. Boilingpoints 588
45. Latent heats of vaporization 590
46. Latent heats of fusion 591
47. Properties of saturated steam 592
48. Properties of superheated steam 596
49. Properties of saturated ammonia vapor 603
50. Properties of saturated carbon dioxide vapor 618
51. Relation between pressure, temperature and per cent NH 3 in solution 628
52. Values of partial pressure of ammonia and water vapors for various temperatures
and per cents of ammonia in solution 632
53. Absorption of gases by liquids 634
54. Absorption of air in water 635
55. Heats of combustion of fuel elements and chemical compounds 636
56. Internal thermal conductivity 639
57. Relative thermal conductivity 642
58. General classification of fuels 648
59. Comparison of cellulose and average wood compositions 650
60. Classification of coals by composition 652
61. Classification of coals by gas and coke qualities 654
62. Composition of peats 655
63. Composition of Austrian lignites 656
64. Composition of English coking coals 658
65. Wilkesbarre anthracite coal sizes and average ash content 659
66. Density and calorific power of natural gas 673
67. Products of wood distillation 676
68. Products of peat distillation 678
69. Products of bituminous coal distillation 680
70. Gas yield of English cannel coals 682
71. Comparison of coke oven and retort coal gas 682
72. Relation between oxygen in coal and hydrocarbon in gas 684
73. Density and calorific power of coke oven gas 684
74. Average distillation products of crude mineral oils 686
75. American mineral oil products 687
76. U.S. gasolene and kerosene bearing crude oils 688
77. Calorific power of gasolenes and kerosenes 691
78. Properties of oilgas 693
79. Yield of retort oil gas 694
80. Density and calorific power of oil gas 694
81. Boudouard's equilibrium relations for CO and CO2 with temperature 697
82. Change of O 2 in air to CO and CO 2 at 1472 F 699
83. Composition of hypothetical air gas, general 704
84. Composition of hypothetical air gas, no CO 2 and no CO 705
85. Density and calorific power of blast furnace gas 708
86. Water gas characteristics with bed temperature 710
87. Composition of hypothetical water gas, general 714
88. Composition of hypothetical water gas, no CO 2 and no CO . 715
LIST OF TABLES xvii
*O. PAGE
89. Density and calorific power of water gas 718
90. Composition of hypothetical producer gas from fixed carbon 725
91. Density and calorific power of producer gas 737
92. Characteristics of explosive mixtures of oil gas and air 747
93. Calculated ignition temperatures for producer gas 761
94. Compressions commonly used in gas engines 762
95. Ignition temperatures 763
96. Variation of ignition temperature of charcoal with distillation temperature 763
97. Per cent detonating mixture at explosive limits of proportion 764
98. Velocity of detonating or explosive waves 766
99. Rate of propagation, feet per second, for different proportions in gaseous mixtures . 768
.00. Rates of combustion for coal 769
lOl. Constants of proportion for rate of coal combustion for use in Eq. (848) 771
.02. Boiler efficiency summaries 799
103. Three examples of heat balance for boilers 800
'.04. Composition and calorific power of characteristic coals 818
.05. Combustible and volatile of coals, lignites and peats 826
i. Paraffines from Pennsylvania petroleums 835
i.07. Calorific power of mineral oils by calorimeter and calculation by density formula
of Sherman and Kropff 836
108. Properties of mineral oils 838
109. Composition of natural gases 841
110. Composition of coke oven and retort coal gas 842
111. Composition of U. S. coke 846
112. Fractionation tests of kerosenes and petroleums 847
113. Fractionation tests of gasolenes 851
114. Composition of blastfurnace gas and air gas 853
LI 5. Rate of formation of CO from CO 2 and carbon 855
L16. Composition of water gas 857
117. Composition of producer gas 858
118. Gas producer tests 864
119. Composition of oil producer gas 866
120. Composition of powdered coal producer gas 866
121. Calorific powers of best airgas mixtures 867
122. Composition of boilerflue gases 868
123. Limits of proportions of explosive airgas mixtures 869
!124. Rate of combustion of coal with draft 870
125. Rate of combustion of coal 871
126. Values of s for adiabatic expansion of steam 912
1 127. Values of s for adiabatic expansion of steam determined from initial and final volumes
only 913
I.L28 1042
129. Diagram factors for Otto cycle gas engines 1046
130. Mechanical efficiencies of gas engines 1050
: 131. Allowable compression for gas engines 1050
1 132. Mean effective pressure factors for Otto cycle engines 1053
jL33. Giildner's values of Otto engine real volumetric efficiency with estimated mean
suction resistances 1055
1 134. Comparative heat balances of gas producer and engine plants 1057
j 135. Heat balances of gas producer plants 1060
JL36. Heat balances of gas and oil engines 1060
137. Steam plant heat balances 1063
xviii LIST OF TABLES
NO. PA<
138. Efficiency factors for reciprocating steam engines and turbines 10(
139. Steam turbine efficiency and efficiency factors with varying vacuum and with
steam approximately at constant initial pressure 10'
140. Efficiency factors for lowpressure steam in piston engines 10'
141. Coefficient of discharge for various air pressure and diameters of orifice (Durley) . IK
142. Values of C for air flow (Weisbach) Ill
143. Flow change resistance factors FR (Reitschel) 11'
TABLE OF SYMBOLS
A =area in square feet.
= constant, in formula for most economical load of a steam engine, Chapter III.
= constant, in pipe flow formula^ Chapter VI.
= excess air per pound of coal, Chapter V.
= pounds of ammonia dissolved per pound of weak liquor, Chapter IV.
a = area in square inches.
= coefficient of linear expansion, Chapter IV.
= constant in equation for the ratio of cylinder sizes for equal work distribution in com
pound engine, Chapter III.
= constant in equation for change in intrinsic energy, Chapter VI.
= constant in equation for specific heat, Chapter IV.
= cubic feet of air per cubic foot of gas in explosive mixtures, Chapter' V.
= effective area of piston, square inches, Chapter I.
B = constant in equation for the most economical load of the steam engine, Chapter III.
= constant in equation for flow in pipes, Chapter VI.
Be. = Baume.
B.H.P. = brake horsepower, Chapters III and VI.
= boiler horsepower, Chapter V.
B.T.U. = British thermal unit.
6 = constant in equation for change in intrinsic energy, Chapter VI.
= constant in equation for specific heat, Chapter IV.
(bk.pr.) =back pressure in pounds per square inch.
C = Centigrade.
= circumference or perimeter of ducts in equations for flow, Chapter VI.
= constant.
= heat suppression factor, Chapter V.
= ratio of pressure after compression to that before compression in gas engine cycles,
Chapter VI.
= specific heat, Chapter IV.
Cc per cent of ammonia in weak liquor, Chapter VI.
C p = specific heat at constant pressure.
C/z=per cent of ammonia in rich liquor, Chapter VI.
C s = specific heat of water, Chapter VI.
Cv specific heat at constant volume.
Cl = clearance expressed in cubic feet.
c = clearance expressed as a fraction of the displacement
= constant,
cu.ft. = cubic foot,
cu. in. = cubic inch.
D = constant in equations for pipe flow, Chapter VI.
= density, Chapter IV.
= diameter of pipe in feet, Chapter VI.
= displacement in cubic feet.
D s = specific displacement, Chapter I.
xix
xx TABLE OF SYMBOLS
d = constant in equation for change in intrinsic energy, Chapter VI.
= diameter of a cylinder in inches, Chapter I.
= diameter of pipe in inches, Chapter VI.
= differential,
(del.pr.) = delivery pressure in pounds per square inch, Chapter II.
E = constant in equation for pipe flow, Chapter VI.
= external latent heat, Chapter IV.
= thermal efficiency, Chapter VI.
EB = thermal efficiency referred to brake horsepower, Chapter III.
Efy = boiler efficiency, Chapter V.
Ef = furnace efficiency, Chapter V.
EI = thermal efficiency referred to indicated horsepower, Chapter III.
E m = mechanical efficiency, Chapter III.
E s = heating surface efficiency, Chapter V.
EV= volumetric efficiency (apparent), Chapter VI.
EV' = volumetric efficiency (true), Chapter VI.
e = as a subscript to log to designate base e.
= constant in equation for change in intrinsic energy, Chapter VI.
ei= ratio of true volumetric efficiency to hypothetical, Chapter II.
62 = ratio of true volumetric efficiency to apparent, Chapter III.
e z = ratio of true indicated horsepower to hypothetical, Chapter II.
F = constant in equation for pipe flow, Chapter VI.
= diagram factor for gas engine indicator cards, Chapter VI.
= Fahrenheit.
= force in pounds.
F F = friction factor, F/?Xvelocity head = loss due to friction, Chapter VI.
FR = resistance factor, FRX velocity head=loss due to resistances, Chapter VI.
Fs=special resistances to flow in equations for chimney draft, Chapter VI.
/= constant in equation for changes in intrinsic energy, Chapter VI.
= function.
ft.=foot.
ft.lb. = footpound.
G = constant in equation for pipe flow, Chapter VI.
= weight of gas per hour in equation for chimney flow, Chapter VI.
G m = maximum weight of gases in equation for chimney flow, Chapter VI.
G. S. = grate surface.
g = acceleration due to gravity, 32.2 (approx.) feet per second, per second.
H = as a subscript to denote high pressure cylinder.
= heat per pound of dry saturated vapor above 32 F.
= heat per cubic foot gas.
= heat transmitted, Chapter IV.
= height of column of hot gases in feet, Chapter VI.
= pressure or head in feet of fluid, Chapter VI.
HA = difference in pressure on two sides of an orifice in feet of air, Chapter VI.
Ho = equivalent head of hot gases, Chapter VI.
H M = pressure in feet of mercury, Chapter VI.
H w = pressure in feet of water, Chapter VI.
H.P. = high pressure.
= horsepower, Chapter I.
TABLE OF SYMBOLS xxi
H.S. = heating surface.
H.P.cap.)=bigh pressure cylinder capacity, Chapter III.
h = heat of superheat.
Ji M = difference in pressure on two sides of an orifice in inches of mercury, Chapter VI.
hw = difference in pressure on two sides of an orifice in inches of water, Chapter VI.
7 = as a subscript to denote intermediate cylinder, Chapter III.
I.H.P. = indicated horsepower.
in. = inch,
in.pr.) = initial pressure in pounds per square inch.
7= Joule's equivalent = 778 (approx.) footpounds per B.T.U.
K = coefficient of thermal conductivity, Chapter IV.
= constant.
= proportionality coefficient in equation for draft, Chapter VI.
K e = engine constant =~~ in expression for horsepower. Chapter III.
oo()00
L=as a subscript to denote lowpressure cylinder.
= distance in feet.
= latent heat, Chapters IV and VI.
= length of stroke in feet, Chapter I.
L=per cent of heat in fuel lost in furnace, Chapter V.
L.P. =low pressure.
(L. P. Cap. )= lowpressure capacity, Chapter II.
1 = constant, Chapter III.
= length, Chapter IV.
Ib. = pound.
log = logarithm to the base 10.
log e = logarithm to the base e.
M = mass.
(M.E.P.) =mean effective pressure, pounds per square foot.
m = constant, Chapter III.
area
= mean hydraulic radius = .
perimeter
= molecular weight, Chapter IV.
= ratio of initial pressure to that end of expansion in Otto and Langen gas cycle,
Chapter VI.
(m.b.p.) =mean back pressure in pounds per square inch,
(m.e.p.) =mean effective pressure in pounds per square inch,
(m.f.p.) =mean forward pressure in pounds per square inch.
N = constant, Chapter III.
= revolutions per minute.
n = cycles per minute.
= constant, Chapter III.
= cubic foot of neutral per cubic foot of gaseous mixture, Chapter V.
= number of degrees exposed on thermometer stem, Chapter IV.
= ratio of volume after expansion to volume before in Atkinson gas cycle, Chapter IV.
= specific volume of dry saturated steam, Chapter VI.
xxii TABLE OF SYMBOLS
O= volume of receiver of compound engine in cubic feet, Chapter III.
P = draft in pounds per square foot, Chapter VI.
= load in kilowatts, Chapter III.
= pressure in pounds per square foot.
PF = static pressure in pounds per square foot lost in wall friction, Chapter VI.
PR= static pressure in pounds per square foot lost in changes of crosssection, etc.,
Chapter VI.
PV= velocity head in pounds per square foot.
p= pressure in pounds per square inch.
p c = mean exhaust pressure, Chapter VI.
p s = mean suction pressure, Chapter VI.
pv= partial pressure of water vapor in air, Chapter VI.
Q = quantity of heat or energy in B.T.U. gained or lost by a body passing from one state to
another.
Q/=heat added from fire in Stirling and Ericsson cycles, Chapter VI.
Qi"=heat added from regenerator in Stirling and Ericsson cycles, Chapter VI.
Q2 /= heat abstracted by water jacket in Stirling and Ericsson cycles, Chapter VI.
Q 2 "=heat abstracted by regenerator in Stirling and Ericsson cycles, Chapter VI.
q = quantity of heat per pound of liquid above 32 F.
R ratio of heating surface to grate surface, Chapter V.
' =gas constant.
* RC^ ratio of cylinder sizes in twostage air compressor or compound engine, Chapters II
and III.
RH= ratio of expansion in highpressure cylinder, Chapter III.
RL= ratio of expansion in lowpressure cylinder, Chapter III.
JRp=ratio of initial to back pressure, Chapters III and VI.
R p = ratio of delivery to supply pressure, Chapter II.
Rv = ratio of larger volume to smaller volume.
r=rate of flame propagation in explosive mixtures, Chapter V.
r/>=pressure differences (maximum minimum) in gas cycles, Chapter VI.
rv= volume differences (maximum minimum) in gas cycles, Chapter VI.
(rec.pr.) = receiver pressure in pounds per square inch, Chapter III.
(rel.pr.) = release pressure in pounds per square inch, Chapter III.
S =per cent of ammonia in solution, Chapter IV.
= piston speed, Chapter I.
= pounds of steam per pound of air in producer blast, Chapter V.
= specific heat, Chapter IV.
= specific heat of superheated steam, Chapter VI.
(Sup.Vol.) = volume of steam supplied to the cylinder per stroke, Chapter III.
s = general exponent of V in expansion or compression of gases.
sp.gr. = specific gravity,
sp.ht. = specific heat,
sq.ft. = square foot,
sq.in. = square inch,
(sup.pr.) = supply pressure, in pounds per square inch.
T = temperature, degrees absolute.
TC = temperature of air, Chapter VI.
TH = temperature of gases in chimney, Chapter VI.
J= temperature in degrees scale.
TABLE OF SYMBOLS xxiii
7= rate of heat transfer in B.T.U. per square foot per hour per degree difference in tem
perature, Chapter IV.
7 = intrinsic energy, Chapter VI.
u = velocity in feet per second.
m = velocity in feet per minute, Chapter VI.
V= volume in cubic feet.
7 A = cubic feet per pound air, Chapter VI.
'a = cubic feet per pound, gas, Chapter VI.
7 L = volume of liquid in cubic feet per pound.
r s = volume of solid in cubic feet per pound.
V = volume of vapor in cubic feet per pound.
v= volume, Chapter IV.
W = work in footpounds.
V.R. = water rate.
w = pounds of water per pound of ammonia in solution, Chapter IV.
= weight in pounds.
WR= pounds of rich liquor per pound of ammonia, Chapter VI.
= compression in the steam engine as a fraction of the stroke, Chapter III.
heat added
= H , Chapter VI.
temperature at beginning of addition X specific heat at constant volume
x = constant in the expression for missing water, Chapter III.
= fraction of liquid made from solid or vapor made from liquid, Chapter VI.
= per cent of carbon burned to CO 2 , Chapter V.
= per cent of nozzle reheat, Chapter VI.
= per cent of steam remaining in highpressure cylinder of compound engine at any point
of the exhaust stroke, Chapter III.
= quantity of heat added in generator of absorption system in addition to the amount of
heat of absorption of 1 Ib. of ammonia, Chapter VI.
ratio of lowpressure admission volume to highpressure admission volume, Chapter III.
Y = total steam used per hour by an engine, Chapter III.
heat added
= 1 + , Chapter VI.
temperature at beginning of addition X specific heat at constant pressure
j/ = per cent of vane reheat Chapter VI.
= ratio of the volume of receiver to that of the highpressure cylinder of the compound
engine, Chapter III.
Z= fraction of the stroke of the steam engine completed at cutoff, Chapter III.
heat added from regenerator
= 1+ , Chapter VI.
temperature at beginning of addition X specific heat at constant volume
'= hypothetical best value of Z.
heat added from regenerator
, Chapter VI.
temperature at beginning of addition X specific heat at constant pressure
2= ratio of R. P.M. to cycles per minute.
ce = an angle, Chapter I.
= coefficient of cubical expansion, Chapter III.
= constant in the equation for latent heat, Chapter VI.
xxiv TABLE OF SYMBOLS
= constant in equation for variable specific heat at constant volume, Chapter VI.
a' = constant in equation for variable specific heat at constant pressure, Chapter VI.
= constant in equation for latent heat, Chapter VI.
= fraction of fuel heat available for raising temperature, Chapter V.
Y = constant in equation for latent heat, Chapter VI.
= ratio of crosssection to perimeter, Chapter IV.
sp. ht. at const, press.
= special value for s for adiabatic expansion or compression =
sp. ht. at const, vol.
' = ratio of specific heat at constant pressure to specific heat at constant volume when
is a variable, Chapter VI.
A = increment.
8 = density in pounds per cubic foot.
,= density in cold gases in equations for chimney draft, Chapter VI.
$H = density of hot gases in equations for chimney draft, Chapter VI.
= coefficient of friction, Chapter VI.
[i = material coefficient in heat transfer expression, Chapter IV.
p= internal thermal resistance, Chapter IV.
S= summation.
a = surf ace thermal resistance, Chapter IV.
u = time in seconds.
$ = entropy, Chapter VI.
<J> = entropy, Chapter VI.
NOTE. A small letter when used as a subscript to a capital in general refers to a po
on a diagram, e.g., P a designates pressure at the point A. Two small letters used as si
scripts together, refer in general to a quantity between two points, e.g., W a b designa
work done from point A to point B.
ENGINEERING THERMODYNAMICS
CHAPTER I
WORK AND POWER. GENERAL PRINCIPLES.
1. Work Defined. Work, in the popular sense of performance of any labor,
is not a sufficiently precise term for use in computations, but the analytical
mechanics has given a technical meaning to the word which is definite and which
is adopted in all thermodynamic analysis. The mechanical definition of work is
mathematical inasmuch as work is always a product of forces opposing motion
and distance swept through, the force entering with the product being limited
to that acting in the direction of the motion. The unit of distance in the
English system is the foot, and of force the pound, so that the unit of work is
the footpound. In the metric system the distance unit is the meter and the
force unit the kilogramme, making the work unit the kilogrammeter. Thus,
the lifting of one pound weight one foot requires the expenditure of one foot
pound of work, and the falling of one pound through one foot will perform one
footpound of work. It is not only by lifting and falling weights that work is
expended or done; for if any piece of mechanism be moved through a distance
of one foot, whether in a straight or curved path, and its movement be resisted
by a force of one pound, there will be performed one footpound of work against
the resistance. It is frequently necessary to transform work from one sys
tem of units to the other, in which case the factors given at the end of this
Chapter are useful.
Work is used in the negative as well as in the positive sense, as the force
considered resists or produces the motion, and there may be both positive and
negative work done at the same time; similar distinctions may be drawn with
reference to the place or location of the point of application of tne force. Con
sider, for example, the piston rod of a directacting pump in which a certain
force acting on the steam end causes motion against some less or equal force
acting at the water end. Then the work at the steam end of the pump may be
considered to be positive and at the water end negative, so far as the move
ment of the rod is concerned; when, however, this same movement causes a
movement of the water, work done at the water end (although negative with
reference to the rod motion, since it opposes that motion) is positive with refer
2 ENGINEERING THERMODYNAMICS
ence to the water, since it causes this motion. It may also be said that the
steam does work on the steam end of the rod and the water end of the rod does
work on the water, so that one end receives and the other delivers work, the rod 
acting as a transmitter or that the work performed at the steam end is the input
and that at the water end the output work.
(See the end of Chapter I for Tables I, II, III, IV, and VI, Units of
Distance, of Surface, of Volume, of Weight and Force, and of Work.)
Example. An elevator weighing 2000 Ibs. is raised 80 ft. How much work is done in
footpounds?
Footpounds = force X distance
=2000 X80 = 160,000 ft.lbs.
Ans. 160,000 ft.lbs.
Prob. 1. A pump lifts 150 gallons of water to a height of 250 ft. How much work does
it do? i
Prob. 2. By means of a jack a piece of machinery weighing 10 tons is raised f in. What
is the work done?
Prob. 3. A rifle bullet weighing 2 oz. travels vertically upward lj miles. What work was
done in footpounds?
Prob. 4. A cubic foot of water falls 50 ft. in reaching a waterwheel. How much work can
it do?
Prob. 5. A piston of an elevator is 12 ins. in diameter and has acting on it a pressure of
80 Ibs. per square inch. What work is done per foot of travel?
Prob. 6. It has been found that a horse can exert 75 Ibs. pull when going 7 miles per
hour. How much work can be done per minute?
Prob. 7. How much work is done by an engine which raises a 10ton casting 50 ft.?
Prob. 8. The pressure of the air on front of a train is 50 Ibs. per square foot when the
speed is 50 miles per hour. If the train presents an area of 50 sq.ft., what work is done in
overcoming wind resistance?
Prob. 9. The pressure in a 10inch gun during the time of firing is 2000 Ibs. per square
inch. How much work is done in ejecting the projectile if the gun is 33 ft. 4 ins. long?
2. Power Defined. Power is defined as the rate of working or the work
done in a given time interval, thus introducing a third unit of mechanics, time,
so that power will always be expressed as a quotient, the numerator being a prod
uct of force and distance, and the denominator time. This is in opposition
to the popular use of the word, which is very hazy, but is most often applied to
the capability of performing much work] or the exertion of great force, thus,
popularly, a powerful man is one who is strong, but in the technical sense a man
would be powerful only when he could do much work continuously and rapidly.
An engine has large power when it can perform against resistance many foot
pounds per minute. The unit of power in the English system is the horsepower,
or the performance of 550 footpounds per second or 33,000 footpounds per
minute, or 1,980,000 footpounds per hour. In the metric system the horse
power is termed chevalvapeur, and is the performance of 75 killogrammeters
= 5422 footpounds per second, or 4500 kilogrammeters = 32,549 footpounds
WORK AND POWER. 3
3er minute, or 270,000 kilogrammeters = 1,952,932 footpounds per hour.
Fable VII at the end of Chapter I gives conversion factors for power units.
Example. The piston of a steam engine travels 600 ft. per minute and the mean force
)f steam acting upon it is 65,000 Ibs. What is the horsepower?
Horsepower = footpounds per minute
oo,UUU
^, distance
force XT
time
33,000
65,000X600
~~
Prob. 1. The drawbar pull of a locomotive is 3000 Ibs. when the train is traveling 50
miles per hour. What horsepower is being developed?
Prob. 2. A mine cage weighing 2 tons is lifted up a 2000ft. shaft in 40 seconds. What
horsepower will be required if the weight of the cable is neglected?
Prob. 3. By direct pull on a cable it is found possible to lift 4 tons 20 ft. per second. With
a differential pulley 40 tons may be lifted 3 ft. per second. What is the difference in power
required?
Prob. 4. A horse exerts a pull of 100 Ibs. on a load. How fast must the load be moved to
develop one horsepower?
Prob. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 Ibs. What
horsepower must be available to maintain this speed? (One knot is a speed of one nautical
mile per hour.)
Prob. 6. It is estimated that 100,000 cu. ft. of water go over a fall 60 ft. high every
second. What horsepower is going to waste?
Prob. 7. The force acting on a piston of a pump is 80,000 Ibs. If the piston speed is 150
ft. per minute, what is the horsepower?
Prob. 8. To draw a set of plows 2j miles per hour requires a drawbar pull of 10,000
Ibs. What must be the horsepower of a tractor, to accomplish this?
Prob. 9. The horsepower to draw a car up a grade is the sum of the power necessary to
pull it on a level and that necessary to lift it vertically the same number of feet as it rises on
the grade. What will be the horsepower required to draw a car 20 miles per hour up a 12 per
cent grade if the car weighs 2500 Ibs. and the drawbar pull on the level is 250 Ibs.?
3. Work in Terms of Pressure and Volume. Another of the definitions
of mechanics fixes pressure as force per unit area so that pressure is always a
quotient, the numerator being force and the denominator area, or length to
the second power. If, therefore, the pressure of a fluid be known, and accord
ing to hydromechanics it acts equally and normally over all surface in contact
with it, then the force acting in a given direction against any surface will be
the product of the pressure and the projected area of the surface, the projection
being on a plane at right angles to the direction considered. In the case of pis
tons and plungers the line of direction is the axis of the cylinder, and the pro
jected area is the area of the piston less the area of any rod passing completely
through the fluid that may be so placed. When this plane area moves in a
4 ENGINEEEING THERMODYNAMICS
direction perpendicular to itself, the product of its area and the distance will be
the volume swept through, and if a piston be involved the volume is technically
the displacement of the piston. Accordingly, work may be expressed in three
ways, as follows:
Work = force X distance ;
Work = pressure X area X distance ;
Work = pressure X volume.
The product should always be in footpounds, but will be, only when appro
priate units are chosen for the factors. These necessary factors are given as
follows :
Work in footpounds = force in Ibs. X distance in ft.
= pressure in Ibs. per sq.ft. X area in sq.ft. X distance in ft
= pressure in Ibs. per sq.in. X area in sq.in. X distance in ft
= pressure in Ibs. per sq.ft. X volume in cu.ft.
= pressure in Ibs. per sq.in. X 144 X volume in cu.ft.
As pressures are in practice expressed in terms not only as above, but also
in heights of columns of common fluids and in atmospheres, both in English and
metric systems, it is convenient for calculation to set down factors of equivalence
as in Table V, at the end of the Chapter.
In thermodynamic computations the pressure volume product as an expres
sion for work is most useful, as the substances used are always vapors and gases,
which, as will be explained later in more detail, have the valuable property of
changing volume indefinitely with or without change of pressure according
to the mode of treatment. Every such increase of volume gives, as a conse
quence, some work, since the pressure never reaches zero, so that to derive work
from vapors and gases they are treated in such a way as will allow them to change
volume considerably with as much pressure acting as possible.
It should be noted that true pressures are always absolute, that is, measured
above a perfect vacuum or counted from zero, while most pressure gages and
other devices for measuring pressure, such as indicators, give results measured
above or below atmospheric pressure, or as commonly stated, above or below
atmosphere. In all problems involving work of gases and vapors, the absolute
values of the pressures must be used; hence, if a gage or indicator measure
ment is being considered, the pressure of the atmosphere found by means of the
barometer must be added to the pressure above atmosphere in order to obtain the
absolute or true pressures. When the pressures are below atmosphere the
combination with the barometric reading will depend on the record. If a record
be taken by an indicator it will be in pounds per square inch below atmosphere
and must be subtracted from the barometric equivalent in the same units to
give the absolute pressure in pounds per square inch. When, however, a
vacuum gage reads in inches of mercury below atmosphere, as such gages
do, the difference between its reading and the barometric gives the absolute
WORK AND POWER. 5
essure in inches of mercury directly, which can be converted to the desired
aits 'by the proper factors.
While it is true that the barometer is continually fluctuating at every place,
frequently happens that standards for various altitudes enter into calculations,
id to facilitate such work, values are given for the standard barometer at various
titudes with equivalent pressures in pounds per square inch in Table IX.
Frequently in practice, pressures are given without a definite statement
: what units are used. Such a custom frequently leads to ambiguity, but it
often possible to interpret them correctly from a knowledge of the nature of the
roblem in hand. For instance, steam pressures stated by a man in ordinary prac
ce as being 100 Ibs. may mean 100 Ibs. per square inch gage (above atmosphere),
ut may be 100 Ibs. per square inch absolute. Steam pressures are then most
Dirimonly stated per square inch and should be designated as either gage or abso
ite. Pressures of compressed air are commonly expressed in the same units as
team, either gage or absolute, though sometimes in atmospheres. Steam pressures
elow atmosphere may be stated as a vacuum of so many inches of mercury,
leaning that the pressure is less than atmosphere by that amount, or may
e given as a pressure of so many inches of mercury absolute, or as so many
ounds per square inch absolute. The pressures of gases stored in tanks under
igh pressure are frequently recorded in atmospheres, due to the convenience
I computation of quantities on this basis. Pressures of air obtained by blowers
>r fans are usually given by the manufacturers of such apparatus in ounces
>er square inch above (or below) atmosphere. Such pressures and also differ
nces of pressure of air due to chimney draft, or forced draft, and the pressure
f illuminating gas in city mains, are commonly stated in inches of water, each
nch of water being equivalent to 5.196 Ibs. per square foot. The pressure of
vater in city mains or other pressure pipes may be stated either in pounds per
iquare inch or in feet of water head.
Example. A piston on which the mean pressure is 60 Ibs. per square inch sweeps through
i volume of 300 cu.ft. What is the work done?
W =PXV, where F=cu.ft. and P=lbs. per sq.ft.
/. TF=60X144X300 =2,592,000 ft.lbs.
Prob. 1. The mean pressure acting per square inch when a mass of air changes in
volume from 10 cu.ft. to 50 cu.ft. is 40 Ibs. per square inch. How much work is done?
Prob. 2. An engine is required to develop 30 H.P. If the volume swept through per
ninute is 150 cu.ft., what must the mean pressure be?
Prob. 3. The mean effective pressure in compressing air from one to five atmospheres is
28.7 Ibs. per square inch. How many horsepower are required to compress 1000 cu.ft. of
free air per minute?
Prob. 4. At an altitude of 4100 ft. a pressure gage showed the pressure on one side of
l piston to be 50 Ibs. per square inch while the pressure on the opposite side is 3 Ibs. per
square inch absolute. What pressure was tending to move the piston?
Prob. 6. At an altitude of 1 mile the mean pressure in a gas engine cylinder during the
suction stroke was found to be 12 Ibs. per square inch absolute. What work was done
by the engine to draw in a charge if the cylinder was 5 ins. in diameter and the stroke 6 ins.?
6 ENGINEERING THERMODYNAMICS
Prob. 6. After explosion the piston of the above engine was forced out so that the ga
volume was five times that at the beginning of the stroke. What must the M.E.P. hav<:
been to get 20,000 ft.lbs. of work?
Prob. 7. On entering a heating oven cold air expands to twice its volume. Wha
work is done per cubic foot of air?
Prob. 8. A projectile is forced from a gun by a constant air pressure of 1000 Ibs. pe
square inch. Before it begins to move there is J a cu.ft. of air in the barrel, and at the instant
it leaves the barrel the volume is 10 cu.ft. What work was done on the projectile?
Prob. 9. Water is forced from a tank against a head of 75 ft. by filling the tank witlj
compressed air. How much work is done in emptying a tank containing 1000 cu.ft.?
4. Work of Acceleration and Resultant Velocity. When a force acting
on a mass is opposed by an equal resistance there may be no motion at all, oil
there may be motion of constant velocity. Any differences, however, bet weeri
the two opposing forces will cause a change of velocity so long as the difference]
lasts, and this difference between the two forces may be itself considered as thcj
only active force. Observations on unresisted falling bodies show that the$
increase in velocity 32.16 ft. per second for each second they are free to fall
and this quantity is universally denoted by g. If then, a body have anj
velocity, ui, and be acted on by a force equal to its own weight in the direction
of its motion for a time, T seconds, it will have a velocity U2 after that time.
It may be that the force acting is not equal to the weight of the body, in which
case the acceleration will be different and so also the final velocity, due to the
action of the force, but the force producing any acceleration will be to the
weight of the body as the actual acceleration is to the gravitational acceleration,
So that
Actual acceleration force _ actual acceleration
Weight of body or gravitational force gravitational acceleration (g) f
and
Actual accelerating force =  r  . ~.  TT X actual acceleration.
gravitational acceleration (g)
or
change of velocity
F= MX. . . . ........... (2)
The work performed in accelerating a body is the product of the resistance
met into the distance covered, L., while the resistance, or the abovedefined force,
acts, or while the velocity is being increased. This distance is the product of
the time of action and the mean velocity, or the distance in feet,
.......... (3)
WORK AND POWER. 7
The work is the product of Eqs. (2) and (3), or, work of acceleration is
Jjr _M(u 2 Ui) ^
where w is the weight in pounds. Exactly the same result will be obtained by
the calculus when the acceleration is variable, so that Eq. (4) is of universal
application.
The work performed in accelerating a body depends on nothing but its mass
and the initial and final velocities, and is in every case equal to the product of
half the mass and the difference between the squares of the initial and final
velocities, or the product of the weight divided by 6^4 and the difference between
the squares of the initial and final velocities.
It frequently happens that the velocity due to the reception of work is desired,
and this is the case with nozzle flow in injectors and turbines, where the steam
performs work upon itself and so acquires a velocity. In all such cases the
velocity due to the reception of the work energy is
64.32TF
U 2 =
where W is work in footpounds and w, as before, is weight in pounds. Or if
the initial velocity be zero, as it frequently is,
U2 = l^= 1 64.32. . . . \" . (6)
\ w \ w
For conversion of velocity units, Table VIII, at the end of the Chapter,
is useful.
Example. A force of 100 Ibs. acts for 5 seconds on a body weighing 10 Ibs.; if the
original velocity of the body was 5 ft. per second, what will be the final velocity, the
distance traveled and the work done?
10 (M.5).
~32^~^5 '
U 2 = 1615 ft. per second;
W= M(U1 * "'> =
405,000 ft.lbs.
Prob. 1. A stone weighing \ Ib. is dropped from a height of 1 mile. With what veloc
ity and in what length of time will it strike if the air resistance is zero?
Prob. 2. A car moving 20 miles per hour and weighing 25 tons is brought to rest in
500 ft. What is the negative acceleration, the time required to stop, and the work done?
8
ENGINEERING THERMODYNAMICS
Prob. 3. Steam escapes through an opening with a velocity of half a mile per second.
How many footpounds of energy were imparted to each pound of it to accomplish this?
Prob. 4. A weight of 100 Ibs. is projected upward with a constant force of 200 Ibs.
How much further will it have gone at the end of 10 seconds than if it had been merely
falling under the influence of gravity for the same period of time?
Prob. 6. A projectile weighing 100 Ibs. is dropped from an aeroplane at the height of
J mile. How soon will it strike, neglecting air resis ance?
Prob. 6. A waterwheel is kept in motion by a jet of water impinging on flat vanes.
The velocity of the vanes is onehalf that of the jet. The jet discharges 1000 Ibs. of
water per minute with a velocity of 200 ft. per second. Assuming no losses, what is
amount of the work done?
Prob. 7. With the wind blowing 30 miles per hour, how much work could a 12ft.
windmill perform if 25 per cent of the available work were utilized.
NOTE. The weight of a cubic foot of air may be taken as .075 Ib.
Prob. 8. An engine has a piston speed of 600 ft. per minute and runs at 150 R.P.M.
If the reciprocating parts weigh 500 Ibs., how much work is done in accelerating the
piston during each stroke?
Prob. 9. A flywheel with rim 10 ft. in diameter to center of section and weighing 5
tons, revolves at a rate of 150 R.P.M. ; 100,000 ft.lbs. of work are expended on it. How
much will the speed change?
5. Graphical Representation of Work. As work is always a product of
force and distance or pressure and volume, it may be graphically expressed by
S 3
D
A 12345
Distances in Feet
FIG. 1. Constant Force, Work Diagram, ForceDistance Coordinates.
an area on a diagram having as coordinates the factors of the product. It is
customary in such representations to use the horizontal distances for volumes
and the vertical for pressures, which, if laid off to appropriate scale and
in proper units, will give footpounds of work directly by the area enclosed.
Thus in Fig. 1, if a force of 5 Ibs. (AB) act through a distance of 5 ft. (BC)
there will be performed 25 footpounds of work as indicated by the area of the
WOEK AND POWER
B
p
D
A 1 2 3 i 5
rectangle ABCD, which encloses 25 unit rectangles, each representing one foot
pound of work.
If a steam cylinder piston suffers a displacement of 5 cu.ft. under the steam
pressure (absolute) of 5 Ibs. per square foot then the operation which results
in the performance of 25 footpounds of work is represented by the diagram
Fig. 2, ABCD.
a
1 3
Volumes iu Cubic Feet
FIG. 2. Constant Pressure Work Diagram, Pressure Volume Coordinates.
Prob. 1. Following the method given for Fig. 1, draw a diagram for the example of
Section 3
Prob. 2. By means of a diagram, show that the work done by a pressure of 1000 Ibs.
per square foot traversing a distance of 10 ft. is 10,000 ft.lbs.
Prob. 3. Draw a diagram for the case of a volume change from 1 to 10 cu.ft. while the
pressure acting is 20 Ibs. per square inch.
Prob. 4. Draw a pressure volume diagram for the case of forcing a piston out of a
cylinder by a water pressure of 15,000 Ibs. per square foot, the volume of the cylinder at
the start is \ cu.ft. and at the end 6 cu.ft. Make a diagram to scale and report work
per square inch of diagram.
Prob. 5. A pump draws in water at a constant suction pressure of 14 Ibs. and dis
charges it at a constant delivery pressure of 150 Ibs. per sq.in. Considering the pump
barrel to be empty at beginning of suction and end of delivery and to contain 3 cu.ft.
when full, draw the diagram for this case and find the footpounds of work done.
Prob. 6. In raising a weight a man pulls on a rope with a constant force of 80 Ibs.
If the weight is lifted 40 ft., find from a diagram the work done.
Prob. 7. In working a windlass a force of 100 Ibs. is applied at the end of a 6ft.
lever, the drum of th windlass being 1 ft. in diameter. Draw a work diagram for
work applied and for work done in lifting if there be no loss in the windlass.
Prob. 8. The steam and water pistons of a pump are on the same rod and the area
of the former is twice that of the latter, the stroke being 3 ft. Show by a diagram
that the work done in the two cylinders is the same if losses be neglected.
Prob. 9. An engine exerts a drawbar pull of 8000 Ibs. at speed of 25 miles an hour.
A change in grade occurs and speed increases to 40 miles per hour and the pull decreases
to 5000 Ibs. Show by a diagram the change in horsepower.
10
ENGINEERING THERMODYNAMICS
6. Work by Pressure Volume Change. Suppose that instead of being
constant the pressure were irregular and, being measured at intervals of 1 cu.ft.
displacement, found to be as follows:
Pressure.
Lbs. per Sq.Ft.
Displacement
Volume Cu.Ft.
100
125
1
150
2
100
3
75
4
50
. 5
150
C '
&
.^
\
lm
A'
E
^^
\
8
^^
\
b
^^
B'
\
D
^x.
o
^^s^
e
\^
E
.g
^\
8 50
^^V^
F
&
H
3
1 2 3 4 5 T
Volumes in Cubic Feet
FIG. 3. Work Diagram, PressureVolume Coordinates.
Relations.
Discontinuous Pressure Volume
This condition might be plotted as in Fig. 3, A, B, C, D, E, F, G, H. The
work done .will be the area under the line joining the observation points. In
the absence of exact data on the nature of the pressure variations between the<
two observation points A and B, a variety of assumptions might be made as to
the precise evaluation of this area, as follows:
(a) The pressure may have remained constant at its original value for th(
first cubic foot of displacement, as shown dotted AB,' and then suddenly have
risen to B. In this case the work done for this step would be 100 footpounds 
(6) Immediately after the measurement at A the pressure may have riser!
to A' and remained constant during displacement A' to B, in which case ttuj
work done would be 125 footpounds.
(c) The pressure may have risen regularly along the solid line AB, in whicl
100+125,
case the work area is a trapezoid and has the value
pounds.
XI = 112.5 foot
WORK AND POWER
11
It thus appears that for the exact evaluation of work done by pressure
volume change, continuous data are necessary on the value of pressure with
respect to the volume. If such continuous data, obtained by measurement or
otherwise, be plotted, there will result a continuous line technically termed the
pressurevolume curve for the process. Such a curve for a pressure volume
change starting at 1 cu.ft. and 45 Ibs. per square foot, and ending at 7 cu.ft.,
and 30 Ibs. per square foot, is represented by Fig. 4, A, B, C, D, E.
The work done during this displacement under continuously varying pressure
is likewise the area between the curve and the horizontal axis when pressures are
laid off vertically, and will be in footpounds if the scale of pressures is pounds
per square foot and volumes, cubic feet. Such an irregular area can be divided
into small vertical rectangular strips, each so narrow that the pressure is sensibly
constant, however much it may differ in different strips. The area of the
rectangle is PAF, each having the width AF and the height P, and the work
assures in Pounds per Square Foot
5 g S g 8
cz
D
^
B
^
*>s
/
X
X
*v
A
/
\
x
\
X
***
E
SO 12 34 56 7
Volumes in Cubic Feet
FIG. 4. Work Diagram, PressureVolume Coordinates. Continuous Pressure Volume
; Relations,
ea will be exactly evaluated if the strips are narrow enough to fulfill the
nditions of sensibly constant pressure in any one. This condition is true only
lor infinitely narrow strips having the width dV and height P, so that each has
the area PdV and the whole area or work done is
W= PdV.
(7)
This is the general algebraic expression for work done by any sort of continuous
pressure volume change. It thus appears that whenever there are available
sufficient data to plot a continuous curve representing a pressure volume change,
the work can be found by evaluating the area lying under the curve and bounded
by the curve coordinates and the axis of volumes. The work done may be
found by actual measurement of the area or by algebraic solution of Eq. (7),
which can be integrated only when there is a known algebraic relation between
the pressure and the corresponding volume of the expansive fluid, gas or vapor.
Prob. 1. Draw the diagrams for the following cases : (a) The pressure in a cylinder
12 ins. in diameter was found to vary at different parts of an 18in. stroke as follows:
12
ENGINEERING THERMODYNAMICS
Pressure in Pounds
per Sq.In.
Per Cent of
Stroke.
100
100
10
100
30
100
50
83.3
60
71.5
70
62.5
80
55.5
90
50.0
100
(6) On a gas engine diagram the following pressures were found for parts of stroke.
IN
OUT
V
P
V
P
V
P
0.25cu.ft.
14.7
0.1
45.2
0.13
146.2
0.20 "
19.5
0.102
79.7
0.15
116.7
0.14 "
29.7
0.104
123.2
0.17
95.7
0.10 "
45.2
0.106
157.7
0.19
80.7
0.108
181.7
0.21
68.7
0.11
188.2
0.23
58.7
0.12
166.2
.
Prob. 2. Steam at a pressure of 100 Ibs. per square inch absolute is admitted to a
cylinder containing .1 cu.ft. of steam at the same pressure, until the cylinder contains
1 cu.ft., when the supply valve closes and the volume increases so that the product of
pressure and volume is constant until a pressure of 30 Ibs. is reached. The exhaust
valve is opened, the pressure drops to 10 Ibs. and steam is forced out until the volume
becomes 1 cu.ft., when the exhaust valve closes and the remaining steam decreases in
volume so that product of pressure and volume is constant until the original point
is reached. Draw the pressure volume diagram for this case.
Prob. 3. During an air compressor stroke the pressures and volumes were as
follows :
Volume in
Cu.Ft.
Pressure in Lbs.
Sq.In.
2.0
14.0
1.8
15.5
1.6
17.5
1.4
20.0
1.2
23.3
1.0
28.0
0.8
28.0
0.4
28.0
0.0
28.0
Draw the diagram to a suitable scale to give work area in footpounds directly.
Prob. 4. Draw the diagrams for last two problems of Section 3.
WOEK AND POWEE 13
7. Work of Expansion and Compression. Any given quantity of gas or
vapor confined and not subject to extraordinary thermal changes such as
explosion, will suffer regular pressure changes for each unit of volume change,
or conversely, suffer a regular volume change for each unit of pressure change,
so that pressure change is dependent on volume change and vice versa. When
the volume of a mass of gas or vapor, Vi, is allowed to increase to 2 by the
movement of a piston in a cylinder, the pressure will regularly increase or
decrease from PI to Pz, and experience has shown that no matter what the gas
or vapor or the thermal conditions, if steady, the volumes and pressures will
have the relation for the same mass,.
(8)
or the product of the pressure and s power of the volume of a given mass
will always be the same. The exponent s may have any value, but usually
lies between 1 and 1.5 for conditions met in practice.
I The precise value of s for any given case depends on
(a) The substance.
(6) The thermal conditions surrounding expansion or compression, s being
different if the substance receives heat from, or loses heat to, .external sur
roundings, or neither receives nor loses.
(c) The condition of vapors as to moisture or superheat when vapors are
under treatment.
Some commonly used values of s are given in Table X at the end of this
chapter for various substances subjected to different thermal conditions dur
ing expansion or compression.
Not only does Eq. (8) express the general law of expansion, but it likewise
expresses the law of compression for decreasing volumes in the cylinder with
corresponding rise in pressure. Expansion in a cylinder fitted with a piston
is called balanced expansion because the pressure over the piston area is
balanced by resistance to piston movement and the mass of gas or vapor is
substantially at rest, the work of expansion being imparted to the piston and
resisting mechanism attached to it. On the other hand when the gas or vapor
under pressure passes through a nozzle orifice to a region of lower pressure the
falling pressure is accompanied by increasing volumes as before, but the work
of expansion is imparted not to a piston, because there is none, but to the fluid
itself, accelerating it until a velocity has been acquired as a resultant of the
work energy received. Such expansion is termed free expansion and the law of
P]q. (8) applies as well to free as to balanced expansion. This equation, then,
is of very great value, as it is a convenient basis for computations of the work
done in expansion or compression in cylinders and nozzles of all sorts involv
ing every gas or vapor substance. Some expansion curves for different values
of s are plotted to scale in Fig. 5, and the corresponding compression curves in
Fig. 6, in which
Curve A has the exponent s=
Curved " " s= .5
14
ENGINEERING THERMODYNAMICS
Curve C has the exponent s = 1.0
Curve D
Curved
Curved
Curved
Curved
s = l.l
8=1.2
8 = 1.3
s = 1.4
s = 1.5
21000
1 2 34 5 6 7 8 9 10 11
FIG. 5. Comparison of Expansion Lines having Different Values'of s.
The volume after expansion is given by
so that the final volume depends on the original volume, on the ratio of the two
pressures and on the value of the exponent. Similarly, the pressure after
expansion
WORK AND POWER
15
epends on the original pressure, on the ratio of the two volumes and on
be exponent.
The general equation for the work of expansion or compression can now be
itegrated by means of the Eq. (8), which fixes the relation between pressures
nd volumes. From Eq. (8),
9 10 11 12 13 14 15 16 17 18 1'J 20
Volumes in Cubic Feet
FIG. 6. Comparison of Compression Curves having Different Values of s.
which, substituted in Eq.(7), gives
but as K is a constant,
V s '
W=K
(11)
The integral of Eq. (11) will have two forms:
(1) When s is equal to one, in which case
(2) When s is not equal to one.
P 2 V 2 = K l ;
16 ENGINEERING THERMODYNAMICS
Taking first the case when s is equal to one,
2F
F
Whence
W
(a)
(6)
^ (c)
^lo&g (d)
When s =
(12)
Eqs. (12) are all equal and set down in different forms for convenience ir
computation; in them
T 7 2 = largest volume = initial vol. for compression = final vol. for expansion
P2 = smallest pressure = initial pres. for compression = final pres. for expansion
Fi = smallest volume = final vol. for compression = initial vol. for expansion
PI = largest pressure = final pres. for compression = initial pres. for expansion
These Eqs. (12) all indicate that the work of expansion and compression Q\
this class is dependent only on the ratio of pressures or volumes at the beginning
and end of the process, and the PV product at either beginning or end, this
product being of constant value.
When the exponent s is not equal to one, the equation takes the form,
rv t jy rv 2
= K\ jT=K\ V~ s dV
Jv V s Jv
As s is greater than one, the denominator and exponents will be negative, so
changing the form to secure positive values,
JL/i L\
slVFi' 1 TV" 1 /'
This can be put in a still more convenient form. Multiplying and dividing aj
1 1
or
7 2 ,i FlS i
v 2
WORK AND POWER
ubstituting the value of K = P 2 V 2 s = PiVi s ,
Whence
17
<>
Whens^l,
. (13)
Eqs. (13) gives the work for this class of expansion and compression in terms
pressure ratios and volume ratios and in them
V 2 = largest volume = initial vol. for compression = final vol. for expansion;
P 2 = smallest pressure = initial pres. f or compression = final pres. for expansion ;
Vi = smallest volume = final vol. for compression = initial vol. for expansion;
PI = largest pressure = final pres. for compression = initial pres. for expansion.
The work of expansion or compression of this class is dependent according
to Eqs. (13), upon the ratio of pressures or volumes at beginning and end of the
process, the exponent, and on the pressure volume product appropriately taken.
It should be remembered that for the result to be in footpounds appropriate
units should be used and all pressures taken absolute. Examination of Eqs.
(12) and (13), for the work done by expansion or compression of both classes,
shows that it is dependent on the initial and final values of pressures and volumes
and on the exponent s, which defines the law of variation of pressure with
volume between the initial and final states.
Example 1. Method of calculating Diagram, Figs. 5 and 6. Consider the curve
for which s = 1.4 as typical of the group.
Assumed Data. Vi = 1 .0 cu.ft. PI = 26,000 Ibs.. per square foot.
Then
PiVi s =K = 20,000 Xl 1 ' 4 =20,000.
For any other value of P, V was found from the relation,
rrr'
.715
18 ENGINEERING THERMODYNAMICS
Let P x = 6000,
then V x =
or
.715
log 3.33 = .5224
.715 X. 5224 = .373= log V t .
/. F*=2.36.
A series of points, as shown below, were found, through which the curve was drawn
p
20,000
P '
20,000
log; p 
1 20,000
T log p.
V
18000
1.111
0.0453
0.032
1.08
14000
1.430
0.1553
0.111
1.30
10000
2.000
0.3010
0.214
1.64
6000
3.330
0.5224
0.373
2.36
2000
10.000
1.0000
0.714
5.18
1000
20.000
1.3010
0.930
8.51
Curves for other values of s were similarly drawn. Starting at a common volume
of 20 cu. ft. the compression curves of Fig. 6 were determined by the same methods
Example 2. A pound of air at 32 F. and under atmospheric pressure is compressed
to a pressure of five times the original. What will be the final volume and the wort
done if s = l and if s = 1.4? The volume of 1 Ib. of air at 32 F. and one atmosphere
is 12.4 cu.ft. approx.
Fors = l,
Pi 
PT 5 '
12.4cu.ft.;
Z?=^ = ^4=5, whence
Vl /2 Vl
! =2.48 cu.ft.
=P 2 F 2 log e = 2116xl2.41oge5;
f 2
= 21 16 X 12.4 X 1 .61 = 42,300 footpounds.
For s = 1.4,
7, _ /PA O
il^*^)
5 may be raised to the .71 power by means of logarithms as follows: (5)' 71 is equal to
the number whose logarithm is .71 log 5.
WORK AND POWER 19
Log 5 =.699, .71 X.699 =.4963, and number of which this is the logarithm is 3.13,
icnce,
Fi = F 2 43.13 or Fi=3.96;
2116X12.4 ft _ lbg
The value of W can also be found by any other form of equation (13) such as,
The value of Vi being found as before, the work expression becomes after numerical
substitution
w 10,580X3.96 [ /3.96y 4 ]
.4 ~\12.4/
As the quantity to be raised to the .4 power is less than one, students may find it
easier to use the reciprocal as follows:
/3.96V4
\12.4/
\3.96/
08 ' ' 632
y ,10,580X3.96 (1 _
Prob. 1. Find V l and W for Example 2 if s =1.2 and 1.3.
Prob. 2. If a pound of air were compressed from a pressure of 1 Ib. per square inch
absolute to 15 Ibs. per square inch absolute find Vi and W when s = 1 and 1.4. F 2 = 180
cu.ft. What would be the H.P. to compress 1 Ib. of air per minute?
Prob. 3. Air expands so that s = l. If Pi = 10,000 Ibs. per square foot, Fi = 10 cu.ft.
jind F 2 = 100 cu.ft. and the expansion takes place in 20 seconds, whaj is the H.P. devel
oped?
Prob. 4. 100 cu.ft. of air at atmospheric pressure is compressed in a cylinder to a pres
sure of 8 atmospheres and then expelled against this constant pressure. Find graphically
and by calculation the footpounds of work done for the case where s = 1 and for the case
where s = 1.4.
Prob. 5. At an altitude of 4000 ft., a r is compressed to a pressure of 60 Ibs. per
sq.in. gage. Find .he H.P. required to compress 1000 cu.ft. of free air per minute.
Prob. 6. From the algebraic equation show how much work is done for a volume
change of 1 to 4, provided pressure is originally 1000 Ibs. per square foot when
(a) PW&,
(6) PV=K 2 ,
(c) PV*=K 3 .
20 ENGINEERING THERMODYNAMICS
Prob. 7. A vacuum pump compresses air from 1 Ib. per square inch absolute to 15
Ibs. per square inch absolute and discharges it. An air compressor compresses air from
atmosphere to 15 atmospheres and discharges it. Compare the work done for equal
initial volumes, s = 1 .4.
Prob. 8. For steam expanding according to the saturation law, compare the work
done by 1 Ib. expanding from 150 Ibs. per square inch absolute to 15 Ibs. per square inch
absolute with, the work of the same quantity expanding from 15 Ibs. to 1 Ib. per square
inch absolute.
NOTE. 1 Ib. of steam occupies 3 cu.ft. at 150 Ibs. per square inch absolute.
Prob. 9. Two air compressors of the same size compress air adiabatically from atmos
phere to 100 Ibs. gage and discharge it. One is at sea level, the other at 10,000 ft. ele
vation. Compare the work in the two cases.
8. Values of Exponent s Defining Special Cases of Expansion or Compres
sion. There are three general methods of finding s for the definition of particular
cases of expansion or compression to allow of the solution of numerical problems.
The first is experimental, the second and third thermodynamic. If by measure
ment the pressures and volumes of a series of points on an expansion or com
pression curve, obtained by test with appropriate instruments, for example,
the indicator, be set down in a table and they be compared in pairs, values of
s can be found as follows: Calling the points A, B, C, etc., then,
and
log Pa~\S log Fa = log Pb + S log Vb,
or
(log 76 log 7a) =log Palog Pb,
hence
_logP.logP.,
~logF,logF.
or
(14)
log
\
log ft
According to Eq. (14a), if the difference between the logarithms of the pressures
at B and A be divided by the differences between the logarithms of the volumes
at A and B respectively, the quotient will be s. According to Eq. (146), the
logarithm of the ratio of pressures, B to A , divided by the logarithm of the ratio
of volumes, A to B respectively will also give s. It is interesting to note that
if the logarithms of the pressures be plotted vertically and logarithms of volumes
horizontally as in Fig. 7, then the line AC equal to the intercept on the horizontal
axis represents the difference between the logarithms of volumes or,
WORK AND POWER
21
; nd similarly
lence
 log P a 
CB
= = tana.
CA
>r the slope of the line indicates the value of s. This is a particularly valuable
nethod, as it indicates at a glance the constancy or variability of s, and there
ire many cases of practice where s does vary. Should s be constant the line
Lvill be straight; should it be variable the line will be curved, but can generally
\
LloffVA
LogVe
BC
.0 .2 .4 .6 .8 1.
Log. V
FIG. 7. Graphic Method of Finding s, from Logarithms of Pressures and Volumes.
be divided into parts, each of which is substantially straight and each will
have a different s. It is sometimes most convenient to take only the beginning
and end of the curve and to use the value of s corresponding to these points,
neglecting intermediate values.
A second method for finding s for a given compression or expansion line
by means of areas is indicated in a note in Section 17 of this Chapter that
is omitted here because it depends on formulas not yet derived. It is by this
sort of study of experimental data that most of the valuable values of s have
been obtained. There is, however, another method of finding a value for
s by purely thermodynamic analysis based on certain fundamental hypo
theses, and the value is as useful as the hypotheses are fair or true to the
facts of a particular case.
One of the most common hypotheses of this sort is that the gas or vapor
undergoing expansion or compression shall neither receive any heat from,
nor give up any to bodies external to itself during the process, and such a process
is given the name adiabatic. Whether adiabatic processes are possible in actual
cylinders or nozzles does not affect the analysis with which pure thermody
namics is concerned. By certain mathematical transformations, to be carried
out later, and based on a fundamental thermodynamic proposition, the adia
22 ENGINEERING THERMODYNAMICS
batic hypothesis will lead to a value of s, the use of which gives results valuable
as a basis of reference, and which when compared with an actual case will per
mit of a determination of how far the real case has departed from the adiabatic
condition, and how much heat has been received or lost at any part of the
process. The particular value of s which exists in an adiabatic change is repre
sented by the symbol y.
Another common hypothesis on which another value of s can be derived,
is that gases in expansion or compression shall remain at a constant temperature
thus giving rise to the name isothermal. This is generally confined to gases
and superheated vapors, as it is difficult to conceive of a case of isothermal or con
stant temperature expansion or compression of wet vapors, as will be seen later.]
In the study of vapors, which, it must be understood, may be dry or wet
that is, containing liquid, a common hypothesis is that during the expansion 01
compression they shall remain just barely dry or that they shall receive or lost
just enough heat to keep any vapor from condensing, or but no more thai
sufficient to keep any moisture that tends to form always evaporated. Expan
sion or compression according to this hypothesis is said to follow the saturatior
law, and the substance to remain saturated. It will appear from this therma
analysis later that the value of s for the isothermal hypothesis is the same for al
gases and equal to one, but for the adiabatic hypothesis s=y will have
different value for different substances, though several may have the sarm
value, while for vapors y will be found to be a variable for any one, its valu<
depending not only on the substance, but on the temperatures, pressures anc
wetness.
When gases or vapors are suffered to expand in cylinders and nozzles o:
caused to compress, it is often difficult and sometimes impossible or perhapn
undesirable to avoid interference with the adiabatic conditions for vaporr
and gases, with the isothermal for gases or with the saturation law for vapors
yet the work to be done and the horsepower developed cannot be predictecj
without a known value of s, which for such cases must be found by expori
ence. A frequent cause of interference with these predictions, which shoukj
be noted, is leakage in cylinders, which, of course, causes the mass undo.
treatment to vary.
According to these methods those values of s have been found which arc
given in Table X, at the end of the Chapter. Mixtures of common gases sucli
as constitute natural, producer, blast furnace or illuminating gas, alone o.l
with air or products of combustion, such as used in internal combustion enginos
have values of s that can be calculated from the elementary gases or measurecj
under actual conditions.
All vapors, except those considerably overheated, have variable exponent;
for adiabatic expansion and compression. This fact makes the exact solutioi
of problems of work for wet vapors, expanding or compressing, which form th
bulk of the practical cases, impossible by such methods as have been described
This class of cases can be treated with precision only by strictly therma
methods, to be described later.
WORK AND POWER
23
Prob. 1. By plotting the values for the logarithms of the following pressures and vol
mes, see if the value for s is constant, and if not find the mean value in each case.
V
10
11
12
11
12
(a) GAS ENGINE COMPRESSION
p V p V
45.2 13 32.2 18
39.7 14 29.7 20
35.7 16 24.7 25
(6) GAS ENGINE EXPANSION
p V p V
188.2
166.2
13
15
17
146.2
116.7
65.7
19
21
23
P
21.0
19.5
14.7
P
80.7
68.7
58.7
V
2.242
2.994
4.556
(c) STEAM EXPANSION
p V p
203.3
145.8
89.9
7.338 52.5
12.44 28.8
22.68 14.7
ob. 2. By plotting the values for the logarithms of the volumes and pressures on
he expansion and compression curves of the following cards, find value for s.
Atmosphere
A.tmosi
24
ENGINEERING THERMODYNAMICS
100
140
Atmosphere
Atmosphere
Prob. 3. From the steam tables at the end of Chapter IV. select the pressures ari
volumes for drysaturated steam and find the value of s between
(a) 150 Ibs. per square inch and 1 Ib. per square inch.
(6) 15
Prob. 4. Find for superheated steam at 150 Ibs. per square inch and with 10(
of superheat expanding to 100 Ibs. per square inch without losing any superheat, tl
corresponding value of s, using tabular data.
Prob. 5. From the ammonia table data for drysaturated vapor find the value of
between
(a) 150 Ibs. per square inch and 1 Ib. per square inch.
(6) 15
9. Work Phases and Cycles, Positive and Negative and Net Work. ACCOM
ing to the preceding it is easy to calculate or predict numerically the work
expansion or compression whenever the conditions are sufficiently definite \
permit of the selection of the appropriate s. It very seldom happens, howev*
that the most important processes are single processes or that the work f
expansion or compression is of interest by itself. For example, before expansic
can begin in a steam cylinder steam must be first admitted, and in air coi
pressors air must be drawn in before it can be compressed. Similarly, aft!
expansion in a steam cylinder there must be an expulsion of used vapor befos
another admission and expansion can take place, while in the air compress i 1
after compression the compressed air must be expelled before more can ent?
for treatment. The whole series of operations is a matter of more concei
than any one alone, and must be treated as a whole. The effect can be me 5
easily found by the summation of the separate effects, and this method :
summation will be found of universal application.
The whole series of processes taking place and involving pressure volur;
changes is called a cycle, any one of them a phase. It is apparent that thdi
can be only a limited number of phases so definite as to permit of the matK
WOEK AND POWER 25
atical treatment necessary for prediction of work, but it is equally clear that
there may be a far greater number of combinations of phases constituting
cycles. Before proceeding to analyze the action of steam or gas in a cylinder
it is necessary first to determine on structural, thermal or any other logical
grounds, what series of separate processes will be involved, in what order, and
the pressure volume characteristics of each. Then and then only, can the
cycle as a whole be treated. These phases or separate and characteristic proc
esses affecting the work done or involving pressure volume changes are divisible
into two classes so far as the causes producing them are concerned, the first
: thermal and the second mechanical. It requires no particular knowledge of
thermodynamics to realize that if air be confined in a cylinder with a free piston
and is heated, that the volume will increase while pressure remains constant,
since the piston will move out with the slightest excess of pressure inside over
what is outside. This is a pressure constant, volume increasing, phase, and
. is thermal since it is a heat effect. If an ample supply of steam be available
; from a boiler held at a constant pressure by the manipulation of dampers and
: fires by the fireman and the steam be admitted to a cylinder with a piston,
i! the piston will move out, the pressure remaining constant and volume increas
ing. This is also a pressure constant, volume increasing phase, exactly as before,
but is mechanical because it is due to a transportation of steam from the boiler
to the cylinder, although in another sense it may be considered as thermal if
the boiler, pipe and cylinder be considered as one part during the admission.
A similar constant pressure phase will result when a compressor piston is
forcibly drawn out, slightly reducing the pressure and permitting the outside 
atmosphere to push air in, to follow the piston, and again after compression of
air to a slight excess, the opening of valves to storage tanks or pipe lines having
a constant pressure will allow the air to flow out or be pushed out of the cylinder
at constant pressure. These two constant pressure phases are strictly mechan
ical, as both represent transmission of the mass. If a cylinder contain water
and heat be applied without permitting any piston movement, there will be
a rise of pressure at constant volume, a similar constant volume pressure rise
phase will result from the heating of a contained mass of gas or vapor under the
same circumstances, both of these being strictly thermal.
However much the causes of the various characteristic phases may differ,
the work effects of similar ones is the same and at present only work effects
are under consideration. For example, all constant volume phases do no
work as work cannot be done without change of volume.
The consideration of the strictly thermal phases is one of the principal
problems of thermodynamics, for by this means the relation between the work
done to the heat necessary to produce the phase changes is established, and a
basis laid for determining the ratio of work to heat, or efficiency. For the
present it is sufficient to note that the work effects of any phase will depend
only on the pressure volume changes which characterize it.
Consider a cycle Fig. 8, consisting of (AB) t admission of 2 cu.ft of steam at
a constant pressure of 200,000 Ibs. per square foot, to a cylinder originally
26
ENGINEERING THERMODYNAMICS
containing nothing, followed by (BC), expansion with s = l, to a pressure o
20,000 Ibs. per square' foot; .(CF), constant volume change of pressure, am
(FG), constant pressure exhaust at 10,000 Ibs. per square foot. These opera
tions are plotted to scale. Starting at zero volume, because the cylindei
200000
456 8 5> 10 11 12 13 It 10 16 17 18 1
Volumes in Cubic Feet
FIG. 8. Analysis of Work Diagram for Admission Expansion and Exhaust of Engine without
Clearance.
originally contains nothing, and at a pressure of 200,000 Ibs. per square foot,
the line AB, ending at volume 2 cu.ft., represents admission and the cross
hatched area under AB represents the 400,000 ft.lbs. of work done during
admission. At B the admission ceases by closure of a valve and the 2 cu.ft.
of steam at the original pressure expands with lowering pressure according
to the law
PaV a =
= 200,000 X 2 = 400,000 ft.lbs.,
So that when
= 4 cu.ft., P =  = 100,000 Ibs. per sq.ft.;
= 5 cu.ft.. p = 40Q > OQQ =
o
V= 10 cu.ft., p =
80,000 Ibs. per sq.ft.;
This continues until F=20 at point C, at which time P= =20,000
^0
Ibs. per square foot, and the work done during expansion is the crosshatched
area JBCD under the expansion curve BC, the value of which can be found by
measuring the diagram or by using the formula Eq. (12),
WOEK AND POWER 27
hich on substitution gives
W bc = 400,000 log e 10 = 400,000X2.3;
= 920,000 ft.lbs.
'his completes the stroke and the work for the stroke can be found by addition
f the numerical values,
W ab = 400,000 ft.lbs. ;
W, c = 920,000 ft.lbs.;
W,a,+Wbc = 1,320,000 ft.lbs.
It is often more convenient to find an algebraic expression for the whole,
ich for this case will be,
= 400,000(1 +log e 10) =400,000X3.3 = 1,320,000 ft.lbs.
the return of the piston it encounters a resistance due to a constant pressure
>f 10,000 Ibs. per square inch, opposing its motion; it must, therefore, do work
n the steam in expelling it. Before the return stroke begins, however, the
>ressure drops by the opening of the exhaust valve from the terminal pressure
f the expansion curve to the exhaust or back pressure along the constant volume
ine, CF t of course, doing no work, after which the return stroke begins, the
>ressure volume line being FG and the work of the stroke being represented by
he crosshatched area DFGH,
W fg = PfVf = 10,000 X 20 = 200,000 ft.lbs.
his is negative work, as it is done in opposition to the movement of the piston.
The cycle is completed by admission of steam at constant zero volume, raising
:he pressure along GA. The net work is the difference between the positive
md negative work, or algebraically
= 1,320,000200,000= l,120,000~ft.lbs.
28
ENGINEERING THERMODYNAMICS
Consider now a cycle of an air compressor, Fig. 9. Admission or suctior
is represented by AB, compression by BC, delivery by CD and constant volum*
drop in pressure after delivery by DA. The work of admission is representec
by the area ABFE or algebraically by
Wab = PbVb,
the work of compression by the area FBCG, or algebraically since s = lA b>
15000
7 8 9 10 11 12 13 It 15 16 17 18 19 20
Volumes in Cubic Feet
FIG. 9. Analysis of Work Diagram for Admission Compression and Delivery of Compresso]
without Clearance.
the work of delivery by the area CDEG, or algebraically
The positive work is that assisting the motion of the piston during suction
the area ABFE or algebraically PbVb. The negative work, that in opposi
tion to the motion, is the sum of the compression and delivery work, the arei
FBCDE, or algebraically,
The net work is the difference and is negative, as such a cycle is mainly resistant
and to execute it the piston must be driven with expenditure of work on the
gas. The valus of the net work is,
WOEK AND POWER
29
n expression which will be simplified in the chapter on compressors. This
et work is represented by the area A BCD, which is the area enclosed by the
ycle itself independent of the axes of coordinates.
It might seem from the two examples given as if net work could be
btained without the tedious problem of summation, and this is in a sense true
the cycle is plotted to scale or an algebraic expression be available, but
hese processes are practically equivalent to summation of phase results. It
light also seem that the work area would always be that enclosed by the
ycle, and this is true with a very important limitation, which enters when
he cycle has loops. If, for example, as in Fig. 10, steam admitted A to B,
xpanded along EC to a pressure C, then on opening the exhaust the pressure
astead of falling to the back pressure or exhaust line as in Fig. 8, would here
8 9 10 11 12 13 11 15 16 17 18 19 20
Volumes in Cubic Feet
10. Analysis of Work Diagram for Engine with Overexpansion Negative Work Loop.
ise along CD, as the back pressure is higher than the terminal expansion pres
ure, after which exhaust will take place at constant back pressure along DE.
The forward stroke work is that under AB and BC or ABC EG, the return
troke work is the area DEGH and the net work is
Area ABC EG Area DEGH.
is the area HGECX is common to both terms of the difference, the net work
y be set down as equal to
Are&ABXHCDX.
It may be set down then in general for looped cycles that the net work area
8 the difference between that of the two loops. If, however, the method laid
lown for the treatment of any cycle be adhered to there need not be any dis
inction drawn between ordinary and looped cycles, that is, in finding the work
30
ENGINEERING THERMODYNAMICS
of a cycle divide it into characteristic phases and group them into positive an
negative, find the work for each and take the algebraic sum.
Special cases of cycles and their characteristics for steam compressors an
gas engine cylinders, as well as nozzle expansion, will be taken up later in mor
detail and will constitute the subject matter of the next two chapters.
Example 1. Method of calculating Diagram, Fig. 8.
Va =0 CU.ft.
Assumed data
To obtain point C.
v d =v c "
V e =2 "
5=1
P C F C =P 6 F 6 or
P a =200,000 Ibs. per square foot.
P 6 =Pa
PC =20,000
P f = 10,000
Pe=Pf.
P & F 6 200,000X2 9n
c =p7" 20,000 =20 '
/. F c =20 and P c = 20,000.
Intermediate points B to C are obtained by assuming various pressures an
finding the corresponding volumes as for F c .
Example 2. Method of calculating Diagram, Fig. 9.
Assumed data
F a =0cu.ft.
F 6 =20 "
V d =0 "
To obtain point C,
P a =21 16 Ibs. per square foot.
P & =P
P c = 14,812 "
( P d =P c
or =
.715
~=7, Iog7=.845 ; and .715 X. 845= log (^Y 4 =. 6105,
or
Therefore,
F c =4.02.
F c =4.02, and P c = 14,812.
Intermediate values BtoC may be found by assuming pressures and finding volumes C(
responding as for V c .
Prob. 1. Steam at 150 Ibs. per square inch absolute pressure is admitted into a cyli
der in which the volume is originally zero until the volume is 2 cu.ft., when the valve
closed and expansion begins and continues until the volume is 8 cu.ft., then exhai
valve opens and the pressure falls to 10 Ibs. absolute and steam is entirely swept 01
Draw the diagram and find the net work done.
WORK AND POWEE 31
Prob. 2. A piston moving forward in, a cylinder draws in 10 cu.ft. of C0 2 at a pressure
f .9 of an atmosphere at sea level and then compresses it adiabatically until the pressure
ises to 9 atmospheres and discharges it at constant pressure. Draw the diagram and
jid net work done.
Prob. 3. A cylinder 18 ins. in diameter and 24 ins. piston stroke receives steam at
00 Ibs. per square inch absolute pressure for  of the stroke. It then expands to the
nd of the stroke and is exhausted at atmospheric pressure. Draw the diagram and
nd the H.P. if the engine makes 100 strokes per minute.
Prob. 4. Two compressors without clearance each with a cylinder displacement of
cu.ft. draw in air at 14 Ibs. per square inch absolute and compress it to 80 Ibs. per
quare inch absolute before delivery. Find the difference in H.P. per 1000 cu.ft. of
*ee air per minute if one is compressing isothermally and the other adiabatically.
)raw diagram for each case.
Prob. 5. A quantity of air 5 cu.ft. in volume and at atmospheric pressure is compressed
1 a cylinder by the movement of a piston until the pressure is 50 Ibs. per square inch
age. If the air be heated the pressure will rise, as in an explosion. In this case the
iston remains stationary, while the air is heated until the pressure reaches 200 Ibs. per
}uare inch gage. It then expands adiabatically to the original volume when the
ressure is reduced to atmosphere with no change in volume. Draw the diagram, and
nd the work done.
Prob. 6. The Brayton cycle is one in which gas is compressed adiabatically and then,
y the addition of heat, the gas is made to expand without change of pressure. Adi
batic expansion then follows to original pressure and the cycle ends by decrease in volume
) original amount without change of pressure. Draw such a cycle starting with 5 cu.ft.
: air at atmospheric pressure, compressing to 4 atmospheres, expanding at constant
ressure to 5 cu.ft., expanding adiabatically to original pressure and finally ending at
iginal point. Find also, work done.
Prob. 7. In the Ericsson cycle air is expanded at constant temperature, cooled at
mstant pressure, compressed at constant temperature and receives heat at constant
)lume. Draw a diagram for the case where 5 cu.ft. at atmospheric pressure are com
essed to 1 cu.ft., heated until volume is 8 cu.ft., expanded to atmosphere and then
>oled to original volume. Find the work.
Prob. 8. In the Stirling cycle constant volume heating and cooling replace that at
tnstant pressure in the Ericsson. Draw diagram starting with 5 cu.ft. and atmsopheric
essure compressing to 1 cu.ft. and then after allowing the pressure to double, expand
i original volume and cool to atmosphere. Find the work.
Prob. 9. The Joule cycle consists of adiabatic compression and expansion and con
ant pressure heating and cooling. Assuming data as in last problem draw the
agram and find the work.
Prob. 10. The Carnot cycle consists of isothermal expansion, adiabatic expansion,
)thermal compression and adiabatic compression. Draw the diagram for this cycle
id find the work.
10. Work Determination by Mean Effective Pressure. While the methods
ready described are useful for finding the .work done in footpounds for a defined
cle with known pressure and volume limits, they are not, as a rule, convenient
r the calculation of the work done in a cylinder of given dimensions. As
prk done can always be represented by an area, this area divided by its length
U give its mean height. If the 'area be in footpounds with coordinates
32 ENGINEERING THERMODYNAMICS
pounds per square foot, and cubic feet, then the division of area in footpoundi
by length in cubic feet will give the mean height or the mean pressure ii
pounds per square foot. Again, dividing the work of the cycle into forward
stroke work and backstroke work, or the respective footpound areas divide<
by the length of the diagram in cubic feet, will give the mean forward pressur
and the mean back pressure. The difference between mean forward pressur
and mean back pressure will give the mean effective pressure, or that averagi
pressure which if maintained for one stroke would do the same work as thj
cycle no matter how many strokes the cycle itself may have required for it !
execution, which is very convenient considering the fact that most gas engine)
require four strokes to complete one cycle. The mean effective pressure mai
also be found directly from the enclosed cycle area, taking proper account e
loops, as representative of net work by dividing this net work area by the lengt
of the diagram in appropriate units. This method is especially convenier
when the diagram is drawn to odd scales so that areas do not give footpounc
directly, for no matter what the scale the mean height of the diagram, whe
multiplied by the pressure scale factor, represents the mean effective pressur;
This mean height can always be found in inches for any scale of diagram hi
finding the area of the diagram in square inches and by dividing by the lengti
in inches, and this mean height in inches multiplied by the scale of pressur.
in whatever units may be used will give the mean effective pressure in the san:
units.
Mean pressures, forward, back or effective, are found and used in two genen j
ways; first, algebraically, and second graphically and generally in this ca?
from test records. By the first method, formulas, based on some assumli
laws for the phases, can be found, and the mean effective pressure and its val j
predicted. This permits of the prediction of work that may be done by a giv i
quantity of gas or vapor, or the work per cycle in a cylinder, or finally the hon
power of a machine, of which the cylinder is a part, operating at a given spei
and all without any diagram measurement whatever. By the second methcJ
a diagram of pressures in the cylinder at each point of the stroke can be obtain 1
by the indicator, yielding information on the scale of pressures. The net wcc
area measured in square inches, when divided by the length in inches, ghssl
the mean height in inches, which, multiplied by the pressure scale per injij
of height, gives the mean effective pressure in the same units, which
usually pounds per square inch in practice.
As an example of the algebraic method of prediction, consider the cyj
represented by Fig. 8. The forward work is represented by
Forward work = P b V b (l+\og e ~j ,
the length of the diagram representing the volume swept through in the
formance of this work is V c , hence
Mean forward pressure = .( l + log c ~J.
WORK AND POWER 33
lut PbVb = P c V c by the law of this particular expansion curve, hence
Mean forward pressure = P c (l+log e ^ c j.
js the back pressure is constant its mean value is this constant value, hence
Constant (mean) back pressure = P/.
y subtraction
Mean effective pressure = P C ( l+log e ^J P f
= 3.3P c P /;
= 3.3X20,00010,000;
= 66,00010,000 = 56,000 Ibs. per sq.ft.
The work done in footpounds is the mean effective pressure in pounds per
re foot, multiplied by the displacement in cubic feet.
Tf = 56,OOOX20 = 1,120,000 ft.lbs. as before.
i(i. 11. GasEngine Indicator Card. For Determination of Mean Effective Pressure
without Volume Scale.
! As an example of the determination of mean effective pressure from a test
1 indicator diagram of unknown scale except for pressures, and without axes
p coordinates, consider Fig. 11, which represents a gas engine cycle in four
jrokes, the precise significance of the lines being immaterial now. The
pressure scale is 180 Ibs. per square inch, per inch of height.
By measurement of the areas in square inches it is found that
Large loop area CDEXC =2.6 sq.in.
Small loop area ABXA =0.5 sq.in.
pc Net cycle area =2.1 sq.in.
Length of diagram . =3.5 in.
Mean height of net work cycle = 0.6 in.
Mean effective pressure = 1 20 X. 06 = 72 Ibs. per square inch.
34 ENGINEERING THERMODYNAMICS
It is quite immaterial whether this diagram were obtained from a large <
a small cylinder; no matter what the size, the same diagram might be secure
and truly represent the pressure volume changes therein. If this particul
cylinder happened to have a diameter of 10 ins. and a stroke of 12 ins. tl
work per stroke can be found. The area of the cylinder will be 78.54 sq.in;
hence the average force on the piston is 72 Ibs. per square inch X78.54 sq.ins.
5654.88 Ibs., and the stroke being 1 ft. the work per stroke is 5654.88 ft.ftj
Both of these methods are used in practical work and that one is adopted \
any particular case which will yield results by the least labor.
Prob. 1. An indicator card from an air compressor is found to have an area of 3.1
sq. ins., while the length is 2^ ins. and scale of spring is given as 80 Ibs. per square in!
per inch height. What is m.e.p. and what would be the horsepower if the compress
ran with a piston speed of 250 ft. per minute and had a piston 9 ins. in diameter?
Prob. 2. For the same machine another card was taken with a 60lb. spring and hj
an area of 4.12 sq.ins. How does this compare with first card, the two having the sa:f
length?
Prob. 3. A steam engine having a cylinder 18 ins. in diameter and a stroke of 24 ir
takes in J cu.ft. of steam at 100 Ibs. absolute, allows it to expand and exhausts at atmt
pheric pressure. An indicator card taken from the same engine showed a length oJ
ins., an area of .91 sq.in. when an 80lb. spring is used. How does the actual m.o
compare with the computed?
Prob. 4. Find m.e.p. by the algebraic method of prediction for,
(a) Bray ton cycle;
(6) Carnot cycle;
(c) Stirling cycle;
(d) Ericsson cycle;
(e) Joule cycle.
(See problems following Section 9).
11. Relation of PressureVolume Diagrams to Indicator Cards. 7<
Indicator. When a work cycle or diagram of pressure volume changes is drai
to scale with pressures and volumes as coordinates, it is termed a press n
volume or PV diagram, and may be obtained by plotting point by point ffm
the algebraic expression for the law of each phase or by modifying the indicai
card. The indicator card is that diagram of pressures and stroke obtai :K
by applying the indicator to a cylinder in operation. This instrument cons fa
essentially of a small cylinder in which a finely finished piston moves frm
without appreciable friction, with a spring to oppose its motion, a pencil mechia
ism to record the extent of the motion, and a drum carrying paper which is mo fc
in proportion to the engine piston movement. The indicator cylinder is ofci
at the bottom and fitted with a ground union joint for attachment to the njto
cylinder through a special cock, which when open permits all the varjta
pressures in the main cylinder to act on the indicator piston, and when clcW
to the main cylinder opens the indicator cylinder to the atmosphere. r f
WORK AND POWER 35
upper side of the indicator piston being always open to the atmosphere, its
movement will be the result of the difference between the pressure in the main
cylinder and atmospheric pressure. A helical spring, carefully calibrated and,
therefore, of known scale, is fixed between the indicator piston and open cap or
, head of its cylinder, so that whenever the pressure in the main cylinder exceeds
atmosphere the indicator piston moves toward the open head of the indicator
cylinder, compressing the spring. Pressures in the main cylinder if less than
atmosphere will cause the indicator piston to move the other way, extending
the spring. This compression and extension of the spring is found in the
calibration of the spring to correspond to a definite number of pounds per
square inch above or below atmosphere per inch of spring distortion, so that
: the extent of the piston movement measures the pressure above or below atmos
,phere. A piston rod projects outward through the cylinder cap and moves a
series of levers and links carrying a pencil point, the object of the linkage being
to multiply the piston movement, but in direct proportion, giving a large
P movement to the pencil for a small piston movement. A cylinder drum carry
ing a sheet of paper is pivoted to the cylinder frame so that the pencil move
} ment will draw on the paper a straight line parallel to the axis of the drum, if
drum is stationary, or perpendicular to it if drum rotates and pencil is sta
tionary. The height of such lines then above or below a zero or datum line,
which is the atmospheric line drawn with the cock closed, measures the pressure
of the fluid under study. The springs have scale numbers which give the
pressure, in pounds per square inch per inch of pencil movement. This paper
*'carrying drum is not fixed, but arranged to rotate about its axis, being pulled
out by a cord attached to the piston or some connecting part through a pro
portional reducing motion so as to draw out the cord an amount slightly less
than the circumference of the drum no matter what the piston movement.
After having been thus drawn out a coiled spring inside the drum draws it back
, on the return stroke. By this mechanism it is clear that, due to the combined
movement of the pencil up and down, in proportion to the pressure, and that
S of the drum and paper across the pencil in proportion to the piston movement,
a diagram will be drawn whose ordinates represent pressures above and below
\ atmosphere and abscissae, piston stroke completed at the same time, or dis
placement volume swept through. It must be clearly understood that such
j indicator diagrams or cards do not give the true or absolute pressures nor the
rjtrue volumes of steam or gas in the cylinder, but only the pressures above or
below atmosphere and the changes of volume of the fluid corresponding to the
J piston movement. Of course, if there is no gas or steam in the cylinder at the
rl beginning of the stroke, the true volume of the fluid will be always equal to the
t . displacement, but no such cylinder can be made.
While the indicator card is sufficient for the determination of mean effective
1,5 pressure and work per stroke, its lack of axes of coordinates of pressure and
^volume prevents any study of the laws of its curves. That such study is
 important must be clear, for without it no data or constants such as the exponent
1 8 can be obtained for prediction of results in other similar cases, nor can the
36
ENGINEERING THERMODYNAMICS
presence of leaks be detected, or the gain or loss of heat during the various
processes studied. In short, the most valuable analysis of the operations is
impossible.
To convert the indicator card, which is only a diagram of stroke or displace
ment on which are shown pressures above and below atmosphere into a pres
sure volume diagram, there must first be found (a) the relation of true or abso
lute pressures to gage pressures, which involves the pressure equivalent of
the barometer, and (6) the relation of displacement volumes to true volume?
of vapor or gas present, which involves the clearance or inactive volume of the
cylinder. The conversion of gage to absolute pressures by the barometer
reading has already been explained, Section 3, while the conversion of displace
ment volumes to true fluid volumes is made by adding to the displacement
volume the constant value in the same units of the clearance, which is usually
the result of irregularity of form at the cylinder ends dictated by structural
necessities of valves, and of linear clearance or free distance between the pis
ton at the end of its stroke and the heads of the cylinder to avoid any possii
bility of touching due to wear or looseness of the bearings.
150 21600
FIG. 12. Ammonia Compressor Indicator Card with Coordinates of Pressures and Volumes
Added to Convert it into a Pressure Volume Diagram.
Let ABCD, Fig. 12, represent an indicator card from an ammonia compressor
on which EF is the atmospheric line. The cylinder bore is 14 ins., stroke
22 ins., and the scale of the indicator spring 100, barometer 28 ins., and measured
clearance 32 cu.in. According to Table IX, 28 ins. of mercury corresponds tc
13.753 Ibs. per square inch, and as 100 Ibs. per square inch, according to the spring
scale, corresponds to 1 in. of height on the diagram, 1 Ib. per square inch cor
responds to 0.01 in. of height, or 13.75 Ibs. per square inch atmospheric pres
sure to .137 in. of height. The zero of pressures then on the diagram must!
lie .137 in. below the line EF. Lay off then a line MH, this distance belo^l
EF. This will be the position of the axis of volume coordinates.
Actual measurement of the space in the cylinder with the piston at the enc I
of its stroke gave the clearance volume of 32 cu. ins. As the bore is 141
ins. the piston area is 153.94 sq. ins. which in connection with the stroke
WORK AND POWER 37
of 22 ins. gives a displacement volume of 22X153.94 = 3386.68 cu. ins.
32
Compared with this the clearance volume is .94 per cent of the
displacement. It should be noted here that clearance is generally expressed
in per cent of displacement volume. Just touching the diagram at the ends
drop two lines at right angles to the atmospheric line intersecting the axis of
volumes previously found at G and H . The intercept GH then represents the
displacement, or 3386.68 cu.in. or 1.96 cu.ft. Lay off to the left of G, .0094, or
in round numbers 1/100 of GH, fixing the point M, MG representing the clearance
to scale, and a vertical through M the axis of pressures. The axes of coordinates
are now placed to scale with the diagram but no scale marked thereon. The
pressure scale can be laid off by starting at M and marking off inch points
each representing 100 Ibs. per square inch. Pounds per square foot can also
be marked by a separate scale 144 times as large. As the length of the diagram
is 2.94 ins. and displacement 1.96 cu.ft., 1 in. of horizontal distance corresponds
to .667 cu.ft. or 1 cu.ft to 1.50 ins. of distance. Lay off then from M dis
tances of 1.50 ins. and mark the first 1 cu.ft. and the second 2 cu.ft., dividing
the intervals into fractions. A similar scale of volumes in cubic inches might
also be obtained.
Ity this process any indicator card may be converted into a pressure volume
diagram for study and analysis, but there will always be required the two factors
of true atmospheric pressure to find one axis of coordinates and the clearance
volume to find the other.
Prob. 1. If in cards Nos. 1 and 2, Section 8, the clearances are 5 per cent and 3 per
cent respectively of the displacement, convert the cards to PV diagrams on the same
base to scales of 4 ins. to 1 cu.ft. and 1 in. to 1000 Ibs. per square foot, for cylinders
9j ins. and 14 ins. respectively in diameter and stroke 12 ins.
Prob. 2. Do the same for cards Nos. 3 and 4, if clearances are 7 per cent and 4 per
cent respectively, for cylinders 10 ins. and 17 ins. respectively in diameter and stroke
12 ins.
Prob. 3. Do the same for cards Nos. 7 and 8 if clearances are 12 per cent and 8 per
3ent respectively and cylinders 12 ins. and 19 ins. in diameter and stroke 24 ins.
12. To Find the Clearance. There are two general methods for the find
ing of clearance, the first a direct volumetric measurement of the space itself
by filling with measured liquid and the second a determination by algebraic
Dr graphic means from the location of two points on the expansion or com
pression curves of the indicator card based on an assumed law for the curves.
The first method of direct measurement is the only one that offers even a
oromise of accuracy, but even this is difficult to carry out because of the
tendency of the measuring liquid to leak past piston or valves, which makes
the result too large if the liquid be measured before the filling of the clearance
space and too small if the liquid be measured after filling and drawing off.
There is also a tendency in the latter case for some of the liquid to remain
i inside the space, besides the possibility in all cases of the failure to completely
; fill the space due to air pockets at high places.
.100
A" A'
A A
B"lB"
B"B\B
40
100
10
Displacement
FIG. 13.
40
6 8
Displacement
10
FIG. 14.
A"
A'\A
20
B'
10
1
Displacement
FIG. 15.
WORK AND POWER
39
By the second general method any two points, A and B, on an expansion or
mpression curve, Figs. 13, 14, 15, may be selected and horizontals drawn to
e vertical line indicating the beginning of the stroke. The points A' and B'
e distant from the unlocated axis an amount A'A" = B'B", representing the
3arance.
Let the clearance volume =CL;
" the displacement up to A =D a ',
the displacement up to B = D b ]
the whole displacement = D ;
" s be the exponent in PV S = constant, which defines the law of the curve.
Then in general,
= D a +Cl,
= D b +Cl,
/ i i\ i i
\P a sP b s ) = P b * D b  P a s D a
Cl
hence the clearance in whatever units the displacement may be measured will be
Cl
P b s D b P a s
]  ! *
Cl
id Cl, in per cent of the whole displacement will be
D b /Pa\sD
b_('
D \P h ) D
Clearance as a fraction of displacement = c =
,Pb
len s = 1 this takes the form
Clearance in fraction of displacement = c
D
. (15)
40 ENGINEERING THERMODYNAMICS
To use such an expression it is only necessary to measure off the atmospheij
pressure below the atmospheric line, draw verticals at ends of the diagraj
and use the length of the horizontals and verticals to the points in the formu j
each horizontal representing one D and each vertical a P.
Graphic methods for the location of the axis of pressures, and hence t'l
clearance, depend on the properties of the curves as derived from analytid
For example, when s=l,
which is the equation of the equilateral hyperbola, a fact that gives a com
name to the law, i.e., hyperbolic expansion or hyperbolic compression,
common characteristics of this curve may be used either separately or toget
the proof of which need not be given here, first that the diagonal of the rectan^e
having two opposite corners on the curve when drawn through the other fr?
corners will pass through the origin of coordinates, and second, that the otb
diagonal drawn through two points of the curve and extended to intersect 11
axes of coordinates will have equal intercepts between each point and t
nearest axis cut.
According to the first principle, lay off, Fig. 16, the vacuum line or asfe
of volume XY and selecting any two points A and B, construct the rectanjp
ACBD. Draw the diagonal CDE and erect at E the axis of pressures E 1 9
then will EZ and EY be the axes of coordinates. According to the seco.l
principle, proceed as before to locate the axes of volumes XY and select t
points, A and B, Fig. 17. Draw a straight line through these points, whil
represents the other diagonal of the rectangle ACBD, producing it to intil
sect XY at M and lay off AN = ~BM. Then will the vertical NE be the a;
of pressures. It should be noted that these two graphic methods apply or
when s = 1 ; other methods must be used when s is not equal to 1 .
A method of finding the axis of zero volume is based upon the slope of t
exponential curve,
PF 5 =c.
Differentiation with respect to V gives
or
Ps
whence
Ps
KS)
WORK AND POWER
41
In other words, the true volume at any given point on the known curve
may be found by dividing the product of P and s by the tangent or the slope
of the line at the given point, with the sign changed. This method gives
results dependent for their accuracy upon the determination of the tangent to
the curve, which is sometimes difficult.
70
Atraos
heric
Line
357
Displacement
FIG. 16.
70
50
30
Atmosi
heric Line
5 7
Displacement
FIG. 17.
11
Graphic Methods of Locating the Clearance Line for Logarithmic or Hyperbolic Expansion
and Compression Curves.
The following graphical solution is dependent upon the principle just given,
and while not mathematically exact, gives results so near correct that the
or is not easily measured. The curve ACB, Fig* 18, is first known experi
lentally or otherwise and therefore the value of s, and the axis FV from
rhich pressures are measured is located. Assume that the axis of zero volume,
42
ENGINEERING THERMODYNAMICS
KP, is not known but must be found. Selecting any two convenient points,
A and B, on the curve, complete the rectangle AHBG with sides parallel and
perpendicular to FV. The diagonal HG cuts the curve at C and the horizontal
axis at E. From C drop the perpendicular CD. If now the distance DE
be multiplied by the exponent s, and laid off DK, and the vertical KP erected,
this may be taken as the zero volume axis.
It cannot be too strongly stated that methods for the finding of clearance
or the location of the axes of pressures from the indicator card, much as they
have been used in practice, are inaccurate and practically useless unless it is
positively known beforehand just what value s has, since the assumed value
PI
dV
'!>!
ID
FIG. 18. Graphic Method of Locating the Clearance Line for Exponential Expansion and
Compression Curves.
of s enters into the work, and s for the actual diagram, as already explained.
is affected by the substance, leakage, by moisture or wetness of vapor and by.
all heat interchange or exchange between the gas of vapor and its container.
Prob. 1. If in card No. 6, Section 8, compression follows the law PV S =K
where s = 1.4 and the barometer reads 29.9 ins. of mercury, locate the axes algebraically
and graphically.
Prob. 2. If in card No. 3, Section 8, expansion follows the law PV S =K
where s = 1.37 and the barometer reads 29.8 ins. of mercury, locate the axes algebraically
and graphically.
Prob. 3. If in card No. 5, Section 8, expansion follows the law PV S =K
where s = l and the barometer reads 27.5 ins. of mercury, locate axes algebraically ami
graphically.
WORK AND POWER
43
13. Measurement of Areas of PV Diagrams and Indicator Cards. Areas
pressure volume diagrams or indicator cards must be evaluated for the
termination of work or mean effective pressure, except when calculation by
rmula and hypothesis is possible. There are two general methods applicable
both the indicator card and PV diagram, that of average heights, and the
animeter measure, besides a third approximate but very useful method,
pecially applicable to plotted curves on crosssection paper.
The third method assumes that the diagram may be divided into strips of
: [ual width as in Fig. 19, which is very easily done if the diagram is plotted on
osssection paper. At the end of each strip, a line is drawn perpendicular to
k le axis of the strip, such that the area intercepted inside the figure is apparently
jual to that outside the figure. If this line is correctly located, the area of the
:ct angular strip will equal the area of the strip bounded by the irregular lines.
()
50
40
:
20
10
^**~
=^
\
^^
>
7
\
\
/
/
\
\
I
\
==*=
^^
1 i
1
I I
Volumes
'IG. 19. Approximate Method of Evaluating Areas and Mean Effective Pressures of
Indicator Cards and P.V. Diagrams.
f the entire figure has irregular ends it may be necessary to subdivide one or
>oth ends into strips in the other direction, as is done at the lefthand side of
r ig. 19. The area of the entire figure will be equal to the summation of lengths
>f all such strips, multiplied by the common width. This total length may be
btained by marking on the edge of a strip of paper the successive lengths in
uch a way that the total length of the strip of paper when measured will be the
otal length of the strips.
The mean height will be the total length of such strips divided by the num
)er of stripwidths in the length of the diagram. By a little practice the proper
ocation of the ends of the strips can be made with reasonable accuracy, and
consequently the results of this method will be very nearly correct if care is
Jxercised.
By dividing the diagram into equal parts, usually ten, and finding the length
)f the middle of each strip, an approximation to the mean height of each strip
44
ENGINEERING THERMODYNAMICS
will be obtained; these added together and divided by the number will grJ
the mean height in inches from which the mean effective pressure may be founj
by multiplying by the scale as above, or the area in square inches by multiphj
ing by the length in inches, which can be converted into work by multiplyiD
by the footpounds per square inch of area as fixed by the scales. As the pre
sures usually vary most, near the ends of the diagram a closer approximation
can be made by subdividing the end strips, as is done in Fig. 20, which repri
sents two steam engine indicator cards taken from opposite ends of the sarr
cylinder and superimposed. The two diagrams are divided into ten equ;
spaces and then each end space is subdivided. The mean heights of the sii
divisions are measured and averaged to get the mean height of the whole en
division, or average pressure in this case for the division. The average heigh
of divisions for diagram No. 1 are set down in a column on the left, while tho;
10
30 40 50 GO 70
Displacement in Per Cent of Stroke
80
yo
100
FIG. 20. Simpson's Method for Finding Mean Effective Pressure of Indicator Cards.
for No. 2 are on the right; the sum of each column divided by ten and multiple
by the spring scale gives the whole m.e.p. The heights of No. 1 in inch
marked off continuously on a slip of paper measured a total of 11.16 ins. ai
for No. 2, 10.79 ins., these quantities divided by 10 (number of strips) and muli
plied by the spring scale, 50, gives the m.e.p., as before. This method
often designated as Simpson's rule.
The best and most commonly used method of area evaluation, whether f
work or m.e.p. determination, is the planimeter, a wellknown instrunie
specially designed for direct measurements of area.
14. Indicated Horsepower. Work done by the fluid in a cylinder, because
is most often determined by indicator card measurements, measures the indicat
horsespower, but the term is also applicable to work that would be done by t
execution of a certain cycle of pressure volume changes carried out at a specifi
rate. The mean effective pressure in pounds per square inch, whether of ;
indicator card or PV cycle, when multiplied by piston area in square inchc
WORK AND POWER 45
ives the average force acting on the piston for one stroke, whether the cycle
squired one, two or x strokes for its execution, and this mean force multiplied
y the stroke in feet gives the footpounds of work done by the cycle. Therefore,
m.e.p. = mean effective pressure in pounds per square inch for the cycle
referred to one stroke;
a = effective area of piston in square inches;
L = length of stroke in feet;
n = number of equal cycles completed per minute;
N = number of revolutions per minute;
$ = mean piston speed = 2LN feet per minute;
N
z = number of revolutions to complete one cycle = . Then will the
licated horsepower be given by,
I.H.P.=
33000
(m.e.p. )LaN
33000z
(m.e.p. )aS
33000X22*
(6)
(17)
When there are many working chambers, whether in opposite ends of the
cylinder or in separate cylinders, the indicated horsepower of each should
e found and the sum taken for that of the machine. This is important not
nly because the effective areas are often unequal, as, for example, in opposite
&ds of a doubleacting cylinder with a piston rod passing through one side
inly or with two piston rods or one piston rod and one tail rod of unequal
iameters, but also because unequal valve settings which are most common
[ill cause different pressure volume changes in the various chambers.
It is frequently useful to find the horsepower per pound mean effective pressure,
r hich may be symbolized by K e , and its value given by
_ Lan LaN
~ 33000 = "330002"
ping this constant, which may be tabulated for various values of n, stroke
nd bore, the indicated horsepower is given by two factors, one involving
pflinder dimensions and cyclic speed or machine characteristics, and the other
jae resultant PV characteristic, of the fluid, symbolically,
I.H.P.=12Xm.e.p.).
foese tables of horsepower per pound m.e.p. are usually based on piston speed
than rate of completion of cycles and are, therefore, directly applicable
46 ENGINEERING THERMODYNAMICS
when z = \ or n = 2N, which means that the two cycles are completed in or
revolution, in which case,
and
aS
33000'
whence
T TT r TS f \ (m.e.p.)a$
I.H.P.=g e (m.e.p.) P
Table XI at the end of this chapter gives values of (H.P. per Ib. m.e.p.]
K e for tabulated diameters of piston in inches and piston speeds in feet p<
minute. Tables are frequently given for what is called the engine constan
which is variously defined as either
(a) oonrwv which must be multiplied by m.e.p. Xn to obtain H.P., or
ooUUU
(6) Qon , which must be multiplied by m.e.p. XLXn to obtain H.P.
ooUUU
For an engine which completes two cycles per revolution, this is the same j
multiplying .by m.e.p. X$. Before using such a table of engine constants \
must be known whether it is computed as in (a) or in (6).
Example. A 9 in. Xl2 in. doubleacting steam engine runs at 250 R.P.M. and 1
mean effective pressure is 30 Ibs. What is H.P. per pound m.e.p. and the I.H.P.?
Lan 1X63.6X500
Ke =
33000 33000
I.H.P.=. 9636X30 =28.908.
Prob. 1. A pump has a piston speed of 250 ft. per minute; piston diameter is 24 ;k,
What is the H.P. per pound mean effective pressure?
Prob. 2. A simple singleacting 2cylinder engine has a piston 10 ins. in diameter wji
a 2in. rod and a stroke of 15 ins. It runs with a mean effective pressure of 45 ;i
per square inch at a speed of 220 R.P.M. What is the H.P.?
Prob. 3. A gas engine has one working stroke in every four. If the speed is W
R.P.M. what must be the m.e.p. to give 6 H.P. when the cylinder has a diameter of 6 s.
and a stroke of 12 ins.?
Prob. 4. An air compressor is found to have a mean effective pressure of 50 Ibs. u
the cylinder is double acting and is 12 ins. diameter and 16 ins. stroke, what H.P. B1
be needed to drive it at 80 R.P.M.?
Prob. 5. A gasoline engine has an engine constant (a) of .3. What must Ix 1
m.e.p. to give 25 H.P.?
Prob. 6. A blowing engine has an m.e.p. of 10 Ibs. Its horsepower is 500. VWj
is the H.P. per pound m.e.p.?
WORK AND POWER 47
Prob. 7. Two engines of the same size and speed are so run that one gives twice the
3ower of the other. How will the engine constants and m.e.p. vary?
Prob. 8. From the diagrams following Section 9 what must have been the H.P. per
pound m.e.p. to give 300 H.P. in each case?
Prob. 9. How will the H.P. per pound m.e.p. vary in two engines if the speed of
me is twice that of the other, if the stroke is twice, if the diameter of piston is twice?
15. Effective Horsepower, Brake Horsepower, Friction Horsepower,
Mechanical Efficiency, Efficiency of Transmission, Thermal Efficiency. Work
s done and power developed primarily in the power cylinder of engines, and is
transmitted through the mechanism with friction loss to some point at which
t is utilized. There is frequently a whole train of transmission which may
nvolve transformation of the energy into other forms, but always with some
xosses, including the mechanical friction. For example, a steam cylinder may
Irive the engine mechanism which in turn drives a dynamo, which transforms
nechanical into electrical energy and this is transmitted to a distance over
,vires and used in motors to hoist a cage in a mine or to drive electric cars.
There is mechanical work done at the end of the system and at a certain rate,
50 that there will be a certain useful or effective horsepower output for the
system, which may be compared to the horsepower primarily developed in
:he power cylinders. A similar comparison may be made between the primary
)ower or input and the power left after deducting losses to any intermediate
Doint in the system. For example, the electrical energy per minute delivered
X) the motor, or motor input, is, of course, the output of the transmission line,
^gain, the electrical energy delivered to the line, or electrical transmission
nput, is the same as dj^namo output, and mechanical energy delivered to the
lynamo is identical with engine output. The comparison of these measure
nents of power usually takes one of two forms, and frequently both; first,
i comparison by differences, and second, a comparison by ratios. The ratio
I )f any horsepower measurement in the system to the I.H.P. of the power cylin
i ler is the efficiency of the power system up to that point, the difference between
.he two is the horsepower loss up to that point. It should be noted that,
is both the dynamo and motor transform energy from mechanical to electrical
>r vice versa, the engine mechanism transmits mechanical energy and the
vires electrical energy, the system is made up of parts which have the
unction of (a) transmission without change of form, and (7>) transformation
>f form. The ratio of output to input is always an efficiency, so that the efficiency
>/ the power system is the product of all the efficiencies of transformation and of
ransfer or transmission, and the power loss of the system is the sum of trans
ormation and transmission losses. Some of these efficiencies and losses have
eceived names which are generally accepted and the meaning of which is gen
Tally understood by all, but it is equally important to note that others have no
lames, simply because there are not names enough to go around. In dealing
vith efficiencies and power losses that have accepted names these names may
vith reason be used, but in other cases where names are differently under
stood in different places or where there is no name, accurate description must be
48 ENGINEERING THERMODYNAMICS
relied on. As a matter of fact controversy should be avoided by definition
of the quantity considered, whether descriptive names be used or not.
Effective horsepower is a general term applied to the output of a machine,
or power system, determined by the form of energy output. Thus, for an engine
it is the power that might be absorbed by a friction brake applied to the shaft,
and in this case is universally called Brake Horsepower. The difference between
brake and indicated horsepower of engines is the friction horsepower of the
engine and the ratio of brake to indicated horsepower is the mechanical efficiency
of the engine. For an engine, then, the effective horsepower or useful horse
power is the brake horsepower. When the power cylinders drive in one machine
a pump or an air compressor, the friction horsepower of the machine is the
difference between the indicated horsepower of the power cylinders and that
for the pump or compressor cylinders, and the mechanical efficiency is the
ratio of pump or air cylinder indicated porsepower to indicated horsepower
of the power cylinders. Whether the indicated horsepower of the air or pump
cylinders can be considered a measure of useful output or not is a matter of
difference of opinion. From one point of view the machine may be as considered
built for doing work on water or on air, in which case these horsepowers may
properly be considered as useful output. On the other hand, the power pump!
is more often considered as a machine for moving water, in which case the
useful work is the product of the weight of water moved into its head in feet,
and includes all friction through ports, passages and perhaps even in pipes or
conduits, which the indicated horsepower of the pump cylinder does not include,
especially when leakage or other causes combine to make the pump cylinder
displacement differ from the volume of water actually moved. With compressors
the situation is still more complicated, as the air compressor may be considered
useful only when its discharged compressed air has performed work in a rock
drill, hoist or other form of an engine, in which case all sorts of measures of use
ful output of the compressor may be devised, even, for example, as the purely
hypothetical^ possible work derivable from the subsequent admission and
complete expansion of the compressed air in a separate air engine cylinder.
Too accurate a definition, then, of output and input energy in machines anc
power systems is not possible for avoidance of misunderstanding, which maj.
affect questions both of power losses and efficiency of transmission and trans
formation whether in a power system or single machine. It is interesting t(
note here that not only is the indicated work of the power cylinder always conj
sidered the measure of power input for the system or machine, but, as in thtj
other cases, it is itself an output or result of the action of heat on the vapor o:[
gas and of the cycle of operations carried out. The ratio of the indicated powe [
or cylinder work, to the heat energy both in footpound units, that was expended
on the fluid is the thermal efficiency of the engine referred to indicated horse j
power or the efficiency of heat transformation into work, the analysis of whicl
forms the bulk of the subject matter of Chapter VI. Similarly, the ratio o
any power measurement in the system to the equivalent of the heat supplied
is the thermal efficiency of so much of the system as is included.
WORK AND POWER 49
Example. It has been found that when the indicated horsepower of an engine is
>0, a generator is giving out 700 amperes at 220 volts. At end of a transmis ion line is a
otor using the output of the generator." This motor on test gave out 180 brake horse
3wer. Assuming no losses in the transmission line, what was the efficiency of the
.otor, of the generator, of the engine, and of the system?
Motor efficiency =
746
NOTE: Volts X amperes = watts, and, watts f 746 =H. P.
Engine and generator efficiency = =82.4%.
74o
250
Efficiency of system = =72% or 82.4x87.2 =72%.
Zoo
b. 1. An engine is belted to a pump; the I.H.P. of the engine is 50, of the pump
', and the pump delivers 1200 gallons water per minute against 100ft. head. What is
he efficiency of each part and of the entire system.
Prob. 2. An engine is geared to air compressor. Upon test, the efficiency of the engine
lone, gearing alone and compressor alone were each 80 per cent. When the com
ressor H.P. was 100 what was that of the engine?
Prob. 3 A waterwheel is run by the discharge from a pump. The B.H.P. of wheel
j found to be 20 when the pump is delivering 45 gallons of water per minute at a
.lead of 1000 Ibs. per square inch. The water I.H.P. of the pump is 30 and the
(team I.H.P. is 40. What are the efficiencies of each part of the system and the overall
fficienc} 7 '?
Prob. 4. Perry gives a rule for the brake horsepower of steam engines as being
iqual to .95 I.H.P. 10. On this basis find the mechanical efficiency of a 500H.P.
jngine from 200 to 500 H.P. Show results by a curve withB.H.P. and per cent efficiency
U coordinates.
Prob. 5. Perry gives a rule for the efficiency of an hydraulic line as H = .71 25 where
tf is the useful power of the pump and / is the indicated. Find / for values of H from
1 00 to 300 and plot a curve of results.
Prob. 6. An engine gives one I.H.P. for every 3 Ibs. of coal per hour. One pound
f coal contains 9,500,000 ft.lbs. of energy. What is the thermal efficiency?
Prob. 7. A gas engine has a mechanical efficiency of 70 per cent when delivering
tower to a generator which in turn has an efficiency of 90 per cent. If the engine uses
j.5 cu.ft. of gas per indicated horsepower hour and the gas contains 700,000 ft.lbs. per
; 'iibic foot, what is the net thermal efficiency of the system?
16. Specific Displacement, Quantity of Fluid per Hour or per Minute per
..H.P. It has been shown that the work done in cylinders by pressure volume
Changes of the vapor or gas depends on the mean effective pressure and on the
jlisplacement, or that there is a relation between I.H.P. and displacement.
The quantity of fluid used also depends on the displacement and may be expressed
50 ENGINEERING THERMODYNAMICS
in cubic feet per minute at either the low pressure or high pressure con
dition when the work is done between two definite pressure limits, or in terau
of pounds per minute or hour, which involves the application of fluid densities
to volumes and which eliminates the double expression for the two conditions
of pressure. The displacement per hour per horsepower, termed the specific
displacement, is the basis of computations on the steam consumption of steair
engines, the horsepower per cubic feet of free air per minute for air compressors
the horsepower per ton refrigeration for refrigerating machines and the con
sumption of fuel per hour per horsepower for gas and oil engines. It is
therefore, a quantity of great importance in view of these applications. Apply
ing the symbols already defined to displacement in one direction of one side oJ
a piston
Displacement in cu.ft. per stroke = I/X
144 '
Displacement in cu.ft. per minute = LX
Displacement in cu.ft. per hour =
T ,.
Indicated horsepower
(m.e.p.)LaN
330QO 330 oOg
Whence expressing displacement per hour per I.H.P. or specific displacemen
in one direction for one side of a piston by D S)
1375Qg
^ ^ =
(m.e.p.)LaN 144(m.e.p.) (m.e.p.)*
330002
From Eq. (19) it appears that the specific displacement is equal to zX!3,75(
divided by the mean effective pressure in pounds per square inch.
If two points, A and B, be so located on the indicator card, Fig. 21, as t<
have included between them a fluid transfer phase, either admission to, o
expulsion from the cylinder, then calling B a = pounds per cubic foot or densit;
at point A, and B 6 = pounds per cubic foot or density at point B } the weigh
of fluid present at A is,
(D a +Ct)8 a Ibs.,
and weight of fluid present at B is
(Db^Cl)^b Ibs.,
whence the weight that has changed places or passed in and out per stroke ifj
(Di\Ct)ou (D a iCi)da Ibs. per stroke.
WORK AND POWER
51
both A and B lie on the same horizontal as A and B', d a = d b = d, the
^nsity of fluid at the pressure of measurement, whence the weight of fluid
n sed per stroke, will be
ad the volume per stroke used at density d is
Db'Da CU.ft.,
rhich compared to the displacement is
Db'Da
D '
\A
\A
B'
21. Determination of Consumption of Fluid per Hour per Indicated Horsepower from
the Indicator Card.
Phis is the fraction of the displacement representing the volume of fluid pass
ng through the machine at the selected pressure. Multiplying the specific
lisplacement by this, there results,
Cu.ft. of fluid per hr. at density (d) per I.H.P.= ^ 5 Db '~ Da ,
md
Lbs. of fluid per hr. per I.H.P =
(20)
ore generally, that is, when A and B are not taken at the same pressures
Lbs. fluid per hr. per I.H.P. = ^ 3 J 5 \(D b +Cl)d l>  (D+C7)$.~ . (21)
The particular forms which this may take when applied to special cases will be
examined in the succeeding chapters.
52 ENGINEERING THERMODYNAMICS
Example. An air compressor whose cylinder is 18x24 ins. (18 ins. in diameter and
stroke 2 ft.) runs at 60 R.P.M. and is double acting. The mean effective air pressure
is 50 Ibs. per square inch. What is the specific displacement?
Cu.ft. per hour = ~p = 60 X2 x X120 =25,600.
T IT P _m.e.p.Layi _50x2x254.5xl20 _ Q9
: ~33,000 33,000
Cu.ft. per hour = 25,600 =
I.H.P. = 92.3 =
or by the formula directly,
13,750 13,750 0>7 ,
= =^/o.
m.e.p. 50
Prob. 1. What will be the cubic feet of free air per hour per horsepower delivered
by a 56x72in. blowing engine with 4 per cent clearance and mean effective pressure of
10 Ibs. per square inch?
Prob. 2. An 18x22in. ammonia compressor works with a mean effective pressure
of 45 Ibs. per square inch. What is the weight of NH S per I.H.P. hour if the speed is
50 R.P.M. and compressor is double acting having a volumetric efficiency of 90 per
cent? Use tabular NH 3 densities.
Prob. 3. A steam engine whose cylinder is 9x12 ins. runs at a speed of 300 R.P.M.
and is double acting. If the m.e.p. is 60 Ibs. and the density of steam at end of the
stroke is .03, how many pounds of steam are used per hour per indicated horsepower?
17. Velocity Due to Free Expansion by PV Method. All the cases examined
.tor the work done by PV cycles have been so far applied only to their execution
in cylinders, but the work may be developed in nozzles accelerating the gas
or vapor in free expansion, giving, as a consequence, a high velocity to the fluid.
It was noted that for cylinders many combinations of phases might be found
worthy of consideration as typical of possible actual conditions of practice,
but this is not true of proper nozzle expansion, which has but one cycle, that of
Fig. 22. That this is the cycle in question is seen from the following considera
tions. Consider a definite quantity of the gas or vapor approaching the nozzle
from a source of supply which is capable of maintaining the pressure. It pushes
forward that in front of it and work will be done, ABCD, equal to the admission
of the same substance to a cylinder, so that its approach A B may be considered
as a constant pressure, volume increasing phase for which the energy comes
from the source of supply. This same substance expanding to the lower pres
sure will do the work CBEF; but there will be negative work equivalent to the
pushing away or displacing of an equivalent quantity of fluid at the low pres
sure, or FEGD, making the work cycle A BEG, in which AG is the excess of
WORK AND POWER 53
nitial over back pressure or the effective working pressure, remaining constant
luring approach and lessening regularly during expansion to zero excess at E.
The work done will be from Eq. (13),
D
c
IG. 22. Pressure Volume Diagram for Noz/le Expansion Measuring the Acceleration
Velocity and Horsepower of Jets.
sl
sl
Whence
r 1
54 ENGINEERING THERMODYNAMICS
Assuming the initial velocity to be zero, and the work of Eq. (22) to be don
on 1 lb., the final or resultant velocity will be according to Eq. (6),
u = V%iW
or
(24
This velocity is in feet per second when pressures are in pounds per square foo
and volumes in cubic feet of 1 lb. of substance, and is known as Zeuner
equation for the velocity of a gas or vapor expanding in a nozzle. It i
generally assumed that such expansion, involving as it does very rapid motio
of the fluid past the nozzle, is of the adiabatic sort, as there seems to be n
time for heat exchange between fluid and walls. As already noted, the vain
of s for adiabatic expansion of vapors is not constant, making the correc
solution of problems on vapor flow through orifices practically impossible b
this method of pressure volume analysis, but as will be seen later the therm*
method of solution is exact and comparatively easy.
NOTE. A comparison of Eqs. (22) and (13) and the figures correspondi
will show that the area under the process curve, which is the same as the wo
done during the compression or expansion, if multiplied by s will equal the ar
to the left of the process curve, which in turn represents, as in Fig. 21,
engines, the algebraic sum of admission, complete expansion, and exhaust wo
areas, or as in Fig. 9 for compressors, the algebraic sum of suction, compressi
and delivery work areas. This statement must not be thought to refer to t
work area of such a cycle as that of Fig. 8, where expansion is incomplete, nor
case of overexpansion, Fig. 10.
Example. In Fig. 22 assume the initial pressure at 100 Ibs. per square inch absolu
back pressure at atmosphere, and expansion as being adiabatic. What will be
work per pound of steam and the velocity of the jet, if Vb is 4.36 cu.ft. and s = 1.3
superheated steam?
= 27,206 X. 608 = 16,541 ft.lbs.;
= 1028 ft. per second.
WORK AND POWER
55
Prob, 1. Taking the same pressure range as above, find W and u for adiabatic expan
sion of air, also for isothermal expansion.
Prob. 2. How large must the effective opening of the suction valve be, in an air
compressor 18x24 ins. to allow the cylinder to properly fill if the mean pressuredrop
through the valve is 1 Ib. per square inch and the compressor runs at 80 R.P.M.?
Prob. 3. What must be the diameter of the inlet valve in a gas engine to fill a
cylinder of i cu.ft. capacity if the lift of the valve is f ins., allowing a pressure drop of
1 Ib. per square inch? Engine makes 150 working strokes per minute.
Prob. 4. It has been found from experiment that the velocity of air issuing from a hole
jin plate orifice is 72 per cent of what would be expected from calculation as above when
ithe absolute pressure ratio is 2 to 1, and 65 per cent when the absolute pressure ratio
jis 1^ to 1. What will be the actual velocity for air flowing from a tank to atmosphere
for these pressure ratios?
Prob. 5. CO 2 stored in a tank is allowed to escape through an orifice into the air.
What will be the maximum velocity of the jet if the pressure on the tank be 100 Ibs.
per square inch gage?
NOTE: 1 Ib. C0 2 at pressure of 100 Ibs. per square inch gage occupies 1.15 cu.ft.
Prob. 6. If ammonia gas and hydrogen were allowed to expand from the same pres
ire, how would their maximum velocities compare? Vol. of 1 Ib. of NH 3 at 50 Ibs. per
;square inch gage is 4.5 cu.ft. Vol. of 1 Ib. of H at same pressure is 77.5 cu.ft.
18. Weight of Flow through Nozzles. Applying an area factor to the
velocity equation will give an expression for cubic flow per second which
I becomes weight per second by introducing the factor, density.
Let the area of an orifice at the point of maximum velocity, u, be A sq.ft.,
then will the cubic feet per second efflux be Au. Assume the point of maxi
mum velocity, having area A, to be that part of the nozzle where the pressure
ihas fallen to P e , Fig. 22, and the gas or vapor to have the density B e pounds
per cubic foot. Then will the nozzle flow in pounds per second be
ut the weight per cubic foot is the reciprocal of the cubic feet per pound,
' e , which it has already been assumed, is the final volume, of one pound of the
 fluid. Hence,
uA
w
TV
This may be put in terms of initial gas or vapor conditions for,
V T7\ I
V c V b\ ^r
Tience
tt
uA
uA
56 ENGINEERING THERMODYNAMICS
Substituting in this the value of u from Eq. (24),
""   c  (25)
This weight will be a maximum for a certain value of the pressure ratio, depend
ing on the value of s only, and this value can be found by placing the first dif
ferential coefficient of w with respect to ( ^ ) equal to zero.
dw o.
To accomplish this, rearrange Eq: (25) as follows:
2 8+1
But as the other factors do not enter to effect the result so long as P& does not
vary, w is a maximum when the bracket
s + 1
is a maximum or when
2
or
or
_
But as ( p^) S cannot be equal to zero in practice, then
which gives the condition that w is a maximum when
/PeVV _S + 1
\Pb) 2 '
WORK AND POWER 57
r maximum flow for given initial pressure occurs when
)' ' (26)
For air expanding adiabatically s= 1.407. Maximum flow occurs when
)
 = .528 and for most common values of s it will be between .50 and .60
*6
j?his result is quite remarkable and is verified by experiment reasonably closely.
It shows that, contrary to expectation, the weight of efflux from nozzles will not
Continuously and regularly increase with increasing differences in pressure, but
lor a given initial pressure the weight discharged per second will have reached
Us limit when the final pressure has been diminished to a certain fraction of the
Initial, and any further decrease of the discharge pressure will not increase the
{low through an orifice of a given area.
The subject of flow in nozzles will be treated more completely in
jvhapter VI.
Prob. 1. For the following substances under adiabatic expansion determine the
jressure ratio for maximum flow and find the rate of flow per square inch of orifice under
[his condition when flow is into a vacuum of 10 ins. of mercury with standard barometer:
(a) Carbon dioxide.
(6) Nitrogen.
(c) Hydrogen.
(d) Ammonia.
(e) Dry steam according to saturation law.
19. Horsepower of Nozzles and Jets. Although, strictly speaking, nozzles
[an have no horsepower, the term is applied to the nozzle containing the
i.rifice through which flow occurs and in which a certain amount of work is
j.one per minute in giving to a jet of gas or vapor initially at rest a certain
linal velocity, and amount of kinetic energy. The footpounds of work per
Sound of fluid multiplied by the pounds flowing per second will give the foot
pounds of work developed per second within the nozzle, and this divided by
';50 will give the horsepower developed by the jet, or the nozzle horsepower,
accordingly,
^X^gVx. (a)
(6)
81 , 3
(27)
58 ENGINEERING THERMODYNAMICS
where the expression in the bracket is the work done per pound of substance
The pressures are expressed in pounds per square foot, areas in square feei
and volumes in cubic feet.
Example. A steam turbine operates on wet steam at 100 Ibs. per square inch abso
lute pressure which is expanded adiabatically to atmospheric pressure. What must b<
the area of the nozzles if the turbine is to develop 50 H.P. ideally?
NOTE : 1 cu.ft. of steam at 100 Ibs. = .23 Ib.
_ s_
By Eq. (26), maximum flow occurs when the pressure ratio is ( ) S , or, fo
1.11
this case when the pressure is 100 s (^~rr) =58 Ibs. per square inch absolute. A;
the back pressure is one atmosphere, the flow will not be greater than for the abov
critical pressure. Substituting it in Eq. (25) will give the flow weight w, and using th<
actual back pressure in Eq. (22) will give the work W.
.11
= 110000 ft.lbs. per pound of steam.
= 198A Ibs. of steam per second.
By Eq. (27 a),
^WXffl _ 110000 X198A
550 = 550
Whence
50
Prob. 1. What will be the horsepower per square inch of nozzle for a turbine usin
hot gases if expansion follows law PVs =k, when s = 1.37, the gases being at a pressure c
200 Ibs. per square inch absolute and expanding to atmosphere.
Let the volume per pound at the high pressure be 2 cu.ft.
Prob. 2. What will be the horsepower per square inch of nozzle for the problem
of Section 17?
Prob. 3. Suppose steam to expand according tolawPFs=/b, where s = 1.111, fror
atmosphere to a pressure of 2 Ibs. per sq. inch absolute. How will the area of the or
fice compare with that of the example to give the same horsepower?
NOTE: V b =2QA.
Prob. 4. Suppose steam to be superheated in the case of the example and of the la*
problem, how will this affect the area of nozzle?
NOTE: Let V* =5 and 32 respectively.
Prob. 5. How much work is done per inch of orifice if initial pressure is 100 Ib;
absolute on one side and final 10 Ibs. absolute on other side of a valve through wbic
air is escaping?
WORK AND POWER 59
GENERAL PROBLEMS ON CHAPTER I
1. An air compressor is required to compress 500 cu.ft. of free air per minute to a
pressure of 100 Ibs. per square inch gage; the compressor is direct connected to a steam
engine. The mechanical efficiency of the machine is 80 per cent. What will be the
steam horsepower if compression is (a) isothermal; (b) adiabatic?
2. A mine hoist weighing 10 tons is raised 4500 ft. in one minute. In the first 20
seconds it is accelerated from rest to a speed of 6000 ft. per minute; during the next 30
seconds speed is constant at this value, and during last 10 seconds it is brought to rest.
What will be (a) work of acceleration for each period; (6) work of lift for each period;
(c) total work supplied*by engine; (d) horsepower during constant velocity period?
3. The engine driving the above hoist is driven by compressed air. If air is supplied at
a pressure of 150 Ibs. per square inch gage and is admitted for threequarters of the
'stroke, then expanded adiabatically for the remainder of the stroke and exhausted to the
atmosphere find (a) what must the piston displacement be to lift the hoist, the work of
acceleration being neglected? (6) To what value could the air pressure be reduced if
air were admitted full stroke?
4. It is proposed to substitute an electric motor for the above engine, installing a
waterpower electric plant at a considerable distance. The type of wheel chosen is one in
which a jet of water issuing from a nozzle strikes against a series of revolving buckets.
The available head at the nozzle is 1000 ft. Assuming the efficiency of the motor to be
85 per cent, transmission 80 per cent, generators 90 per cent, and waterwheels 60 per
cent, what will be the cubic feet of water per minute?
5. A steam turbine consists of a series of moving vanes upon which steam jets issuing
from nozzles impinge. It is assumed that for best results the speed of the vanes should be
half that of the jets. The steam expands from 100 Ibs. per square inch gage to 5 Ibs.
per square inch absolute, (a) What must be the best speed of vanes for wet steam where
s = l.lll? (6) If 55 per cent of the work in steam is delivered by the wheel what must
I be the area of nozzle per 100 H.P., and weight of steam per hour 100 H.P?
NOTE: F 6 =3.82.
6. It has been found that a trolley car uses a current of 45 amperes at 550 volts when
li running 12 miles per hour. If the motor is 80 per cent efficient, what is tractive effort?
NOTE : Volts X amperes = watts, and watts f .746 = H.P.
7. An engine whose cylinder is 18x24 ins. (diameter = 18 ins., stroke =24 ins.) is
double acting and runs at a speed of 125 R.P.M. Steam is admitted for onequarter
stroke at a pressure of 125 Ibs. per square inch gage, allowed to expand for the rest of
:i the stroke and then exhausted into a vacuum of 28.25 ins. of Hg. (a) Draw a PV
diagram of the cycle and find the m.e.p. from the diagram and by calculation, and then
find the horsepower. (6) Consider steam to be admitted onehalf stroke without
other change. How will the horsepower vary? (c) What will be the horsepower for
onequarter admission if the exhaust pressure is 15 Ibs. per square inch absolute? (d)
What will be the horsepower if the steam pressure be made 150 Ibs. per square inch
absolute, other conditions as in (a)? (e) Suppose all conditions as in (a) but the
speed lowered to 75 R.P.M. What will be the horsepower?
8. Assuming that 50 per cent of the work in the jet is transformed to useful work,
what must be the total area of the nozzles of a steam turbine to develop the same horse
60 ENGINEERING THERMODYNAMICS
power as the engine in problem (7 a), the pressure range being the same and s being 1.3?
7^=3.18.
9. Water is being pumped from the bottom of a shaft 700 ft. deep at the rate of 1000
gallons per minute by an electrically driven pump. Efficiency of pump is 70 per cent,
motor 90 per cent, transmission line 95 per cent, generator 85 per cent, and mechanical
efficiency of engine 80 per cent. What will be the indicated horsepower of the engine?
If the above installation were replaced with an airdriven pump of 65 per cent efficiency, i
efficiency of transmission being 100 per cent, and that of the compressor and engine:
80 per cent, what would be the horsepower of this engine?
10. Show by a PV diagram, assuming any convenient scales, that the quantity of air
discharged byacompressor and the horsepower, both decrease as the altitude increases, and
that the horsepower per cubic foot of air delivered increases under the same condition.
11. A centrifugal pump is driven by a steam engine directly connected to it. The 1
pump is forcing 1000 gallons of water per minute against a head 'of 250 ft. and runs at a
speed of 450 R.P.M. The engine is double acting and its stroke equals the diameter of
the cylinder. Steam of 100 Ibs. per square inch gage is admitted for half stroke, allowed
to expand the rest of the stroke so that s = 1, and is then exhausted to atmosphere. Whatj
must be the size of the engine if the pump efficiency is 65 per cent and the engine
efficiency 75 per cent?
12. (a) What will be the pounds of steam used by this engine per hour per hoi
power? (6) If the steam were admitted but onequarter of the stroke and the initialj
pressure raised sufficiently to maintain the same horsepower, what would be the new
initial pressure and the new value of the steam used per horsepower per hour?
NOTE: Weight of steam per cubic foot for (a) is .261; for (6) is .365.
13. If it were possible to procure a condenser for the above engine so that the exhaust
pressure could be reduced to 2 Ibs. per square inch absolute, (a) how much would the
power be increased for each of the two initial pressures already given? (6) How would
the steam consumption change?
14. A motorfire engine requires a tractive force of 1300 Ibs. to drive it 30 miles perl
'hour, its rated speed. The efficiency of engine and transmission is 80 per cent. When
the same engine is used to actuate the pumps 70 per cent of its power is expended OD
the water. What will be the rating of the engine in gallons per minute when pumping
against a pressure of 200 Ibs. per square inch?
15. A compressor when compressing air at sea level from atmosphere to 100 Ibs. pel
square inch absolute, expends work on the air at the rate of 200 H.P., the air being com
pressed adiabatically. (a) How many cubic feet of free air are being taken into the
compressor per minute and how many cubic feet of high pressure air discharged?
Compressor is moved to altitude of 8000 ft. (6) What will be the horsepower if the same
amount of air is taken in and how many cubic feet per minute will be discharged?
(c) What will be the horsepower if the same number of cubic feet are discharged as in
case (a) and what will be the number of cubic feet of low pressure air drawn in?
(d) Should superheated ammonia be substituted for air at sea level, what would be the
necessary horsepower?
16. A car weighing 40 short tons is at rest and is struck by a train of four cars, each
of the same weight as the first, (a) Upon impact the single car is coupled to the train and
all move off at a certain velocity. If the original velocity of the train was 3 miles
per hour, what will it be after attachment of the extra car? (b) If instead of coupling,
the extra car after impact moved away from the train at twice the speed the train was
then moving, what would be the speed of train?
WORK AND POWER , 61
17. To drag a block of stone along the ground requires a pull of 1000 Ibs. If it be
placed on rollers the pull will be reduced to 300 Ibs., while if it be placed on a wagon with
veilmade wheels, the pull will be but 200 Ibs. Show by diagram how the work required
;o move it 1000 ft. will vary.
18. 100 cu. ft. of air, at atmospheric pressure, is compressed to six times its
)riginal pressure, (a) What will be the difference in horsepower to do this in 45 seconds
sothermally and adiabatically at an elevation of 8000 ft. (6) What will be the final vol
imes? (c) What will be the difference in horsepower at sea level? (d) What will be the
inal volumes?
19. An engine operating a hoist is run by compressed air at 80 Ibs. per square inch
;age. The air is admitted half stroke, then expanded for the rest of the stroke so that
'=1.3 and then exhausted to atmosphere. The engine must be powerful enough to lift
i, ton cage and bring it up to a speed of 2000 ft. per minute in 15 seconds. What will be
jhe necessary displacement per minute?
20. Construct PV diagrams for Probs. 1, 11, 13 and 15, showing by them that the
jyork of admission, compression or expansion, and discharge or exhaust, is equal to that
jound algebraically.
21 The elongation of wrought iron under a force F is equal to the force times the
3ngth of the piece divided by 25,000,000 times the crosssection of metal in the piece.
L4^ in. pipe has 3.75 sq. ins. of metal; a line of this pipe 1 mile long is running full of
rater with a velocity of 600 ft. per minute. This is stopped in 1 second by closing a
ralve. Assuming pipe did not burst, what would be the elongation?
22. Two steam turbines having nozzles of equal throat areas are operating on a steam
liressure of 150 Ibs. per square inch gage. One is allowing steam to expand to atmosphere
Hie other to 2 Ibs. per square inch absolute, both cases having an exponent for expansion
If 1.11. Find the relation of the horsepower in the two cases.
23. The power from a hydroelectric plant is transmitted some distance and then
ued to drive motors of various sizes. At the time of greatest demand for current it has
seen found that 1000 horsepower is given out by the motors. Taking the average
ijfficienoy of the motors as 70 per cent, transmission efficiency as 85 per cent, generator
Ifficiency as 85 per cent, and waterwheel efficiency as 70 per cent, how many cubic feet
If water per second will the plant require if the fall is 80 ft.?
24. A small engine used for hoisting work is run by compressed air. Air is admitted
)r threequarters of the stroke and then allowed to expand for the rest of the stroke in
jch a way that s = 1 .4 and finally exhausted to atmosphere. For the first part of the
oisting,full pressure (80 Ibs. per square inch gage) is applied, but after theload has been
dcelerated the pressure is reduced to 30 Ibs.. per square inch gage. If the engine has
#o cylinders each 9 Xl2 in., is double acting and runs at 200 R.P.M., what will be the
i) horsepower in each case, (6) the specific displacement?
\ 25. A steam engire has a cylinder 12x18 in., is double acting, and runs at 150
['..P.M. (a) What is the engine constant, and (b) horsepower per pound m.e.p.?
ft 26. A waterpower site has available at all times 3500 cu.ft. of water per min
ffe at a 100ft. fall. Turbines of 70 per cent efficiency are installed which take the place
l two doubleacting steam engines whose mechanical efficiencies were 85 per cent.
fie speed of the engines was 150 R.P.M., m.e.p. 100 Ibs. per square inch, and stroke was
esjvice the diameter. What was the size of each engine?
i! 27. Assuming the frictional losses in a compressor to have been 15 per cent, how many
jkf*ft. of gas per minute could a compressor operated by the above engines compress
BI atmosphere to 80 Ibs. per square inch gage if s = 1 .35?
62
ENGINEERING THERMODYNAMICS
TABLE I
CONVERSION TABLE OF UNITS OF DISTANCE
Meters. 1
Kilometers.
Inches.
Feet.
Statute Miles.
Nautical Miles.
1
0.001
39.37
3.28083
0.000621370
0.000539587
1000
1
39370.1
3280.83
0.62137
0.539587
0.0254
0.0000254
1
0.083333
0.0000157828
0.0000137055
0.304801
0.0003048
12
1
0.000189394
0.000164466
1609.35
1.60935
63360
5280
1.
0.868382
1853.27
1.85327
72963.2
6080.27
1.15157
1.
1 In accordance with U. S. Standards (see Smithsonian Tables).
TABLE II
CONVERSION TABLE OF UNITS OF SURFACE
Sq. Meters.
Sq. Inches.
Sq. Feet.
Sq. Yards.
Acres.
Sq. Miles.
1
.000645
.0929
1550.00
1
144
10.76387
.00694
1
1.19599
.111
.000247
.8361
1296
9
1
.000206
4046 87
43560
4840
1
001562
2589999
27878400
3097600
640
1
TABLE III
CONVERSION TABLE OF UNITS OF VOLUME
Cu. Meters.
Cu. Inches.
Cu. Feet.
Cu. Yards.
Lities
(1000 Cu. Cm.)
Gallons (U.S.)
1
61023.4
1
35.3145
.000578
1.3079
1000
.016387
264.170
00433
.028317
. 76456
1728
46656
1
27
.03704
1
28.317
7 . 4805
201 974
.001
.003785
61.023
231
.035314
. 13368
.001308
.004951
1
3.7854
.26417
1
TABLES
63
TABLE IV.
CONVERSION TABLE OF UNITS OF WEIGHT AND FORCE
Kilogrammes.
Metric Tons.
Pounds.
U. S. or Short Tons.
British or Long Tons.
1.
0.001
2.20462
0.00110231
0.000984205
1000.
1.
2204.62
1 . 10231
0.984205
0.453593
0.000453593
1.
0.0005
0.000446429
907.186
0.907186
2000.
1.
0.892957
1016.05
1.01605
2240.
1.12000
1.
TABLE V
CONVERSION TABLE OF UNITS OF PRESSURE
Pounds per
Square Foot.
Pounds per
Square Inch.
Inches of
Mercury at
32 F.
Atmospheres
(Standard at
Sea Level).
'ie Ib. per sq. ft
1
0.006944
0.014139
0004724
le Ib per sq. in. . .
144
1
2 03594
06802
'ie ounce per sq. in
lie atmosphere (standard at sea
level)
lie kilogramme per square meter . .
*ie gramme per square millimeter .
tie kilogramme per square centi
meter
9.
2116.1
20.4817
204.817
2048 17
0.0625
14.696
0.142234
1.42234
14 2234
0.127246
29 . 924
0.289579
2.89579
28 9579
0.004252
1.
0.009678
0.09678
9678
FLUID PRESSURES
lie ft. of water at 39.1 F. (max.
dens.) ....
62 . 425
0.43350
0.88225
029492
lie ft. of water at 62 F
l(ie in. of water at 62 F. . .
62.355
5.196
0.43302
. 036085
0.88080
. 07340
0.029460
002455
ic in. of mercury at 32 F. (stand
lard) ' . .
70 7290
491174
1.
033416
!>iie centimeter of mercury at C. .
ie ft. of air at 32 F., one atmos.
press
27.8461
08071
0.193376
0005604
0.393701
0011412
0.013158
00003813
lie ft. of air, 62 F
0.07607
0005282
0010755
00003594
I * PRESSURES MEASURED BY THE MERCURY COLUMN. For temperatures other than 32 F., the density
'mercury, pounds per cubic inch, and hence the pressure, pounds per square inch, due to a column of mercury
ich high, is given with sufficient accuracy by the following formula:
p = 0.4912a32) X0.0001.
The mercurial barometer is commonly made with a brass scale which has its standard or correct length
5 F, and a linear coefficient of expansion of about 0.000001 for each degree Fahrenheit. Hence, to
<rect the standard of mercury at 32 F., the corrected reading will be
'ere H t is the observed height at a temperature of t F.
64
ENGINEERING THERMODYNAMICS
TABLE VI
CONVERSION TABLE OF UNITS OF WORK
Kilogrammeters.
Footpounds.
Foot Tons (Short Tons).
Foot Tons (Long Tons).
1.
7.23300
0.00361650
0.00322902
0.138255
1.
0.000500
0.000446429
276.510
2000.
1.
0.892857
309.691
2240.
1 . 12000
1.
See also more complete table of Units of Work and Energy in Chapter IV on Work and Heat.
TABLE VII
CONVERSION TABLE OF UNITS OF POWER
Footpounds per
Second.
Footpounds per
Minute.
Horsepower.
ChevalVapeur.
Kilogrammeters pe
Minute.
1.
60.
0.00181818
0.00184340
8.29531
0.0166667
1.
0.000030303
0.0000307241
. 138252
550.000
33000.
1.
1.01387
4562.42
542.475
32548.5
0.986319
1.
4500.00
0.120550
7.23327
0.000219182
0.000222222
1.
TABLE VIII
UNITS OF VELOCITY
Feet per Minute.
Feet per Seconi
One foot per second . ...
60.
1.
One foot per minute . .
1.
0.016667
One statute mile per hour
88.
1.4667
One nautical mile per hour = 1 knot . .
101.338
1.6890
One kilometer per hour
54 6806
0.911344 
One meter per minute
3.28084
0.054581
One centimeter per second
2.00848
0.032808
TABLES
65
TABLE IX
TABLE OF BAROMETRIC HEIGHTS, ALTITUDES, AND PRESSURES
(Adapted from Smithsonian Tables)
Barometric heights are given in inches and millimeters of mercury at its standard density
(32 F.).
Altitudes are heights above mean sea level in feet, at which this barometric height is
standard. (See Smithsonian Tables for corrections for latitude and temperature.)
Pressures given are the equivalent of the barometric height in Ibs. per sq. in. and per
sq. ft.
Standard Barometer.
Altitude, Feet above
Sea Level.
Pressure, Pounds per
Inches.
Centimeters.
Square Inch.
Square Foot.
17.0
43.18
15379
8.350
1202.3
17.2
43.69
15061
8.448
1216.6
17.4
44.20
14746
8.546
1230.7
17.6
44.70
14435
8.645
1244.8
17.8
45.21
14128
8.742
1259.0
18.0
45.72
13824
8.840
1273.2
18 2
46.23
13523
8.940
1287.3
18.4
46.73
13226
9.038
1301.4
18.6
47.24
12931
9.136
1315.6
18.8
47.75
12640
9.234
1329.7
19.0
48.26
12352
9.332
1343.8
19.2
48.77
12068
9.430
1357.9
19 .4
49.28
11786
9.529
1372.1
19.6
49.78
11507
9.627
1386.3
19.8
50.29
11230
9.726
1400.4
20.0
50.80
10957
9.825
1414.6
20.2
51.31
10686
9.922
1428.7
20 4
51.82
10418
10.020
1442.9
20^6
52.32
10153
10.118
1457.0
20.8
52.83
9890
10.217
1471.2
21.0
53.34
9629
10.315
1485.3
21 .2
53.85
9372
10.414
1499.4
21 4
54.36
9116
10.511
1513.6
21 .6
54.87
8863
10.609
1527.7
21.8
55.37
8612
10.707
1541.8
22.0
55.88
8364
10.806
1556.0
22.2
56.39
8118
10.904
1570.1
22.4
56.90
7874
11.002
1584.3
22 6
57.40
7632
11.100
1598.4
22.8
57.91
7392
11.198
1612.6
23.0
58.42
7155
11.297
1626.7
23.2
58.92
6919
11.395
1640.8
23.4
59.44
6686
11.493
1655.0
23.6
59.95
6454
11.592
1669.3
23.8
60.45
6225
11.690.
1683.3
24.0
60.96
5997
11.788
1697.4
24.2
61.47
5771
11.886
1711.6
24.4
61.98
5547
11.984
1725.7
24.6
62.48
5325
12.083
1739.9
24.8
62.99
5105
12.182
1754.0
25.0
63.50
4886
12.280
1768.2
25.2
64.01
4670
12.377
1782.3
25.4
64.52
4455
12.475
1796.5
25 6
65.02
4241
12.573
1810.7
1 25.8
65.53
4030
12.671
1824.8
66
ENGINEERING THERMODYNAMICS
TABLE IX Continued
Standard Barometer.
Altitude, Feet above
Sea Level.
Pressure, Pounds per
Inches.
Centimeters.
Square Inch.
Square Foot.
26.0
65.04
3820
12.770
1838.9
26.1
66.30
3715
12.819
1846.0
26.2
66.55
3611
12.868
1853.1
26.3
66.80
3508
12.918
1860.2
26.4
67.06
3404
12.967
1867.3
26.5
67.31
3301
13.016
1874.3
26.6
67.57
3199
13.065
1881.4
26.7
67.82
3097
13.113
1888.5
26.8
68.08
.2995
13 163
1895.5
26.9
68.33
2894
13.212
1902.6
27.0
68.58
2793
13.261
1909.7
27.1
68.84
2692
13.310
1916.7
27.2
69.09
2592
13.359
1923.8
27.3
69.34
2493
13.408
1930.9
27.4
69.60
2393
13.457
1938.0
27.5
69.85
2294
13.507
1945.1
27.6
70.10
2195
13.556
1952.1
27.7
70.35
2097
13 . 605
1959.2
27.8
70.61
1999
13.654
1966.3
27.9
70.87
1901
13.704
1973 . 3
28.0
71.12
1804
13.753
1980.4
28.1
71.38
1707
13.802
1987.5
28.2
71.63
1610
13.850
1994.5
28.3
71.88
1514
13.899
2001 . 6
28.4
72.14
1418
13.948
2008.7
28.5
72.39
1322
13.998
2015.7
28.6
72.64
1227
14.047
2022.8
28.7
72.90
1132
14.096
2030.0
28.8
73.15
1038
14.145
2037.0
'28.9
73.40
943
14.194
2044.1
29.0
73.66
849
14.243
2051.2
29.1
73.92
756
14.293
2058.2
29.2
74.16
663
14.342
2065.3
29.3
74.42
570
14.392
2072 . 4
29.4
74.68
477
14.441
2079 . 4
29.5
74.94
384
14.490
2086.5
29.6
75.18
292
14.539
2093.6
29.7
75.44
261
14.588
2100.7
29.8
75.69
109
14.637
2107.7
29.9
75.95
+18
14.686
2114.7
29.92
76.00
14.696
2116.1
30.0
76.20
 73
14.734
2121.7
30.1
76.46
163
14.783
2128.8
30.2
76.71
253
14.833
2135.9
30.3
76.96
343
14.882
2143.0
30.4
77.22
433
14.931
2150.1
30.5
77.47
522
14.980
2157.2
30.6
77.72
611
15.030
2164.2
30.7
77.98
700
15.078
2171.3
30.8
78.23
788
15.127
2178.4
30.9
78.48
877
15.176
2185.5
31.0
78.74
965
15.226
2192.6
TABLES
67
TABLE X
VALUES OF s IN THE EQUATION PV S = CONSTANT FOR VARIOUS SUBSTANCES
AND CONDITIONS
Substance.
s
Remarks or Authority.
All gases
Isothermal
1 1
All gases and vapors. .
All saturated vapors . .
All gases and vapors . .
Air
Constant pressure
Isothermal
Constant volume
Adiaba'ic
00
1.4066
Accepted thermody
narnic law
Smithsonian Tables
Air
Compressed in cylinder
1 4
Experience
iAjnmonia (NH 3 )
Ammonia (NH 3 )
Bromine . . .
Adiabatic, wet
Adiabatic, superheated
Adiabatic
1.1
1.3
1.293
Average
Thermodynamics
Strecker
Carbon dioxide (CO 2 ) .
Carbon monoxide (CO)
Carbon disulphide
(CSz)
Adiabatic
Adiabatic
Adiabatic
1.300
1.403
1.200
Rontgen, Wullner
Cazin, Wullner
Beyne
Chlorine (Cl)
Adiabatic
1.323
Strecker
Chloroform
(CC1 3 CH(OH) 2 )....
Ether (C 2 H 5 OC 2 H 5 ). . .
Hydrogen (H 2 )
Adiabatic
Adiabatic
Adiabatic
1.106
1.029
1 410
Beyne, Wullner
Muller
Cazin
Hydrogen sulph . (H 2 S)
Methane (CH 4 )
Nitrogen (N 2 )
Nitrous oxide (NO 2 ) . .
Pintsch gas
Adiabatic
Adiabatic
Adiabatic
Adiabatic
Adiabatic
1.276
1.316
1.410
1.291
1'.24
Muller
Muller
Cazin
Wullner
Pintsch Co.
Sulphide diox (SO 2 ) ...
Steam, superheated . ..
Steam, wet
Steam wet
Adiabatic
Adiabatic
Adiabatic
Adiabatic
1.26
1.300
Variable
1.111
Cazin, Muller
Smithsonian Tables
(From less than 1 to
more than 1.2)
Rankine ,
Steam, wet
Adiabatic
1+14X% moist.
Perry
Steam, wet
Adiabatic
1.035 + .1X% moist.
Gray
Steam, wet
Steam, dry
Expanding in cylinder
Saturation law
1.
1.0646
Average from practice
Begnault
ENGINEERING THERMODYNAMICS
TABLE XI
HORSEPOWER PER POUND MEAN EFFECTIVE PRESSURE.
aS
VALUE OF K e =
33000
Diameter
of
Cylinder,
Inches.
Speed of Piston in Feet per Minute.
100
200
300
400
500
600
700
800
900
4
0.0381
0.0762
0.1142
0.1523
0.1904
0.2285
0.2666
0.3046
0.3427
4*
0.0482
0.0964
0.1446
0.1928
0.2410
0.2892
0.3374
0.3856
0.4338
5
0.0592
0.1190
0.1785
0.2380
0.2975
0.3570
0.4165
0.4760
0.5355
5*
0.0720
0.1440
0.2160
0.2880
0.3600
0.4320
0.5040
0.5760
0.6480
6
0.0857
0.1714
0.2570
0.3427
0.4284
0.5141
0.5998
0.6854
0.7711
61
0.1006
0.2011
0.3017
0.4022
0.5028
0.6033
0.7039
0.8044
0.9050
7
0.1166
0.2332
0.3499
0.4665
0.5831
0.6997
0.8163
0.9330
1.0490
n
0.1339
0.2678
0.4016
0.5355
0.6694
0.8033
0.9371
1.0710
1.2049
8
0.1523
0.3046
0.4570
0.6093
0.7616
0.9139
1.0662
1.2186
1.3709
8
0.1720
0.2439
0.5159
0.6878
0.8598
1.0317
1.2037
1.3756
1.5476
9
0.1928
0.3856
0.5783
0.7711
0.9639
1 . 1567
1.3495
1.5422
1 . 7350
9*
0.2148
0.4296
0.6444
0.8592
1.0740
1.2888
1.5036
1.7184
1.9532
10
0.2380
0.4760
0.7140
0.9520
1 . 1900
1.4280
1.6660
1.9040
2.1420
11
0.2880
0.5760
0.8639
1.1519
1.4399
1 . 7279
2.0159
2 . 3038
2.5818
12
0.3427
0.6854
1.0282
1.3709
1.7136
2.0563
2.3990
2.7418
3.0845
13
0.4022
0.8044
1.2067
1.6089
2.0111
2.4133
2.8155
3.2178
3.6200
14
0.4665
0.9330
1.3994
1 . 8659
2.3324
2.7989
3.2654
3.7318
4.1983
15
0.5355
1.0710
1.6065
2.1420
2.6775
3.2130
3.7485
4.2840
4.8195
16
0.6093
1.2186
1.8278
2.4371
3.0464
3.6557
4.2650
4.8742
5.4835
17
0.6878
1.2756
1.9635
2.6513
3.3391
4.0269
4.6147
5.4026
6.1904
18
0.7711
1.5422
2.3134
3.0845
3.8556
4.6267
5.3987
6.1690
6.4901
19
0.8592
1.7184
2.5775
3.4367
4.2858
5.1551
6.0143
6.8734
7.7326
20
0.9520
1.9040
2.8560
3.8080
4.7600
5.7120
6.6640
7.6160
8.5680
21
1.0496
2.0992
3.1488
4.1983
5.2475
6.2975
7.3471
8.3966
9.4462
22
.1519
2.3038
3.4558
4.6077
5.7596
6.9115
8.0643
9.2154
10.367
23
.2590
2.5180
3.7771
5.0361
6.2951
7.5541
8.8131
10.072
11.331
24
.3709
2.7418
4.1126
5.4835
6.8544
8.2253
9.5962
10.967
12.338
25
.4875
2.9750
4.4625
5.9500
7.4375
8.9250
10.413
11.900
13.388
26
.6089
3.2178
4.8266
6.4355
8.0444
9.6534
11.262
12.871
14.480
27
.7350
3.4700
5.2051
6.9401
8.6751
10.410
12.145
13.880
15.615
28
.8659
3.7318
5.5978
7.4637
9.3296
11.196
13.061
14.927
16.793 [
29
2.0016
4.0032
6.0047
8.0063
10.008
12.009
14.011
16.013
18.014
30
2.1420
4.2840
6.4260
8.5680
10.710
12.852
14.994
17.136
19.278 [
31
2.2872
4.5744
6.8615
9.1487
11.436
13.723
16.010
18.287
20.585 [
32
2.4371
4.8742
7.3114
9.7485
12.186
14.623
17.060
19.497
21.934 1
33
2.5918
5.1836
7.7755
10.367
12.959
15.551
18.143
20.735
23.326 I
34
2.7513
5.5026
8.2538
11.005
13.756
16.508
19.259
22.010
24.762 I
35
2.9155
5.8310
8.7465
11.662
14.578
17.493
20.409
23.224
26.240 
36
3.0845
6.1690
9.2534
12.338
15.422
18.507
21.591
24.676
27.760 1
37
3.2582
6.5164
9.7747
13.033
16.291
19.549
22.808
26.066
29.324 1
38
3.4367
6.8734
10.310
13 . 747
17.184
20.620
24.057
27.494
30.930 1
39
3.6200
7.2400
10.860
14.480
18.100
21.720
25.340
28.960
32.580
TABLES
TABLE XI Continued
69
Diameter
of
Speed of Piston in Feet per Minute.
Cylinder,
Inches.
100
200
300
400
500
600
700
800
900
40
3.8080
7.6160
11.424
15.232
19.040
22.848
26.656
30.464
34.272
41
4.0008
8.0016
12.002
16.003
20.004
24.005
28.005
32.006
36.007
42
4.1983
8.3866
12.585
16.783
20.982
25.180
29.378
33.577
37.775
43
4.4006
8.8012
13.202
17.602
22.003
26.404
30.804
35.205
39.606
44
4.6077
9.2154
13.823
18.431
23.038
27.646
32.254
36.861
41.469
45
4.8195
9.6390
14.459
19.278
24.098
28.917
33.737
38.556
43.376
46
5.0361
10.072
15.108
20.144
25.180
30.216
35.253
40.289
45.325
47
5.2574
10.515
15.772
21.030
26.287
31.545
36.802
42.059
47.317
48
5.3845
10.967
16.451
21.934
27.418
32.901
38 . 385
43.868
49.352
49
5.7144
11.429
17.143
22.858
28.572
34.286
40.001
45.715
51.429
50
5.9500 11.900
17.850
23.800
29.750
35.700
41.650
47.600
53.550
51
6. 1904  12.381
18.571
24.762
30.952
37.142
43.333
49.523
55.713
52
6.4355
12.871
19.307
25.742
32.178
38.613
45.049
51.484
57.920
53
6.6854
13.371
20.056
26.742
33.427
40.113
46 . 798
53.483
60.169
54
6.9401
13 . 880
20.820
27.760
34.700
41.640
48.581
55.521
62.461
55
7.1995
14.399
21.599
28.798
35.998
43.197
50.397
57.596
64.796
56
7.4637
14.927
22.391
29.855
37.318
44.782
52.246
59.709
67.173
57
7.7326
15.465
23.198
30.930
38.663
46.396
54.128
61.861
69.594
58
8.0063
16.013
24.019
32.025
40.032
48.038
56.044
64.051
72.057
59
8.2849
16.570
24.854
33.139
41.424
49.709
57.993
66.278
74.563
60
8.5680
17.136' 25.704
34.272
42.840
51.408
59.976
68.544
77.112
 1
70
ENGINEERING THERMODYNAMICS
GENERAL FORMULA RELATING TO PRESSURE VOLUME
CALCULATIONS .OF WORK AND POWER
Work = W = Force X Distance =FxL;
= Pressure xArea X Distance =P XA XZ>;
= Pressure X Volume change =Px(V 2 V l ).
Force of acceleration =mass X acceleration.
.
g di
Work of acceleration = mass X difference of (velocity) 2 ,
1 w
__
w
__
,,.
_,*).
Velocity due to work of acceleration = square root of the sum of (initial velocity)
plus 20Xwork per pound of substance accelerated.
w
or if initial velocity is zero,
w
w
Pressure Volume Relation for Expansion or Compression
PV S ^PiVi 8 =P 2 V 2 S =K, a constant. (See Table VIII for values of s.)
log ~
(Note graphical method for finding s when variable, see text.)
FORMULA RELATING TO PRESSURE VOLUME CALCULATIONS 71
Work done during a pressure volume change represented by the equation (PV 8 K)
tween points represented by 1 and 2 in figure =area under curve, Wi.
If 8=1,
rvi C V2 dV
= I PdV=Kl
JVi J Vi V s
= 1.
If s is not equal to 1,
Oi
p.r.r, /PA
^iL^W Jj
Work of admission, complete expansion and exhaust for engines = area to left of
curve (see figure) TF 2 . Same for admission, compression and expulsion for com
pressors. Both cases without clearance.
(When s = l, Wz is same as area under curve, W\ (see above).
Clearance, expressed as a fraction of displacement =c, as a volume
72
ENGINEERING THERMODYNAMICS
Cl D D \P
Indicated horsepower = (mean effective pressure, pounds per square inch X effect! 1
area of piston, square itches X length of stroke, feetXnumber of working cycles pei
formed per minute) divided by 33,000.
T TIP =''
33000
Specific displacement = displacement in one direction for one side of a piston, u
cu.ft. per hour per H.P. =D S = 13,750, z
(m.e.p.)
where z is the number of strokes required o compl e one cycle.
Velocity of a jet due to its own expansion =the square root of the product of 2g Xworl
done by admission, complete expansion and exhaust of 1 Ib. of the substance.
Weight of flow through nozzle or orifices, pounds per second =w = (velocity, feet
second X area, square feet, of orifice) f (volume per pound of substance at section where
area is measured).
1 ( sl )
W= ^A = SQ2 ( ' }' \P V fl^ ^~*~1 f
Maximum discharge w for a given initial pressure occurs when
CHAPTER II
VORK OF COMPRESSORS, HORSEPOWER AND CAPACITY OF AIR, GAS AND
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS.
1. General Description of Structure and Processes. There is quite a
arge class of machines designed to receive a cylinder full of some gas at one
Constant pressure and after the doing of work on the gas through decreasing
Volumes and rising pressures, to discharge the lesser volume of gas against a
Constant higher pressure. These machines are in practice grouped into sub
classes, each having some specific distinguishing characteristic. For example,
plowing engines take in air at atmospheric pressure or as nearly so as the
ralve and port resistance will permit, and after compression deliver the air
it a pressure of about three atmospheres absolute for use in blast furnaces.
These blowing engines are usually very large, work at low but variable speeds,
mt always deliver against comparatively low pressures; they, therefore, have
he characteristics of large but variable capacity and low pressures. A great
^ariety of valves and driving gears are used, generally mechanically moved
uction and automatic spring closed discharge valves, but all valves may be
oitomatic. The compressor cylinder is often termed the blowing tub and the
ompressed or blast air frequently is spoken of as wind by furnace men. They
ire all directconnected machines, an engine forming with the compressor one
nachine. The engine formerly was always of the steam type, but now a change
b being affected to permit the direct internal combustion of the blast furnace
vaste gases in the cylinders of gas engines. These gasdriven blowing engines,
'ihowing approximately twice the economy of steamdriven machines, will in
:ime probably entirely displace steam in steel plants, and this change will take
>lace in proportion to the successful reduction of cost of repairs, increase of
t : eliability and life of the gasdriven blowing engines to equal the steamdriven,
feme lowpressure blowers are built on the rotary plan without reciprocating
bistons, some form of rotating piston being substituted, and these, by reason
f greater leakage possibilities, are adapted only to such low pressures as 5 Ibs.
>er square inch above atmosphere or thereabouts. These blowers are coming
tito favor for blasting gas producers, in which air is forced through thick coal
>eds either by driving the air or by drawing on the gas produced beyond the
>ed. They are also used for forcing illuminating gas in cities through pipes
>therwise too small, especially when the distances are long. In general very
ow pressures and large capacities are the characteristics of the service whether
he work be that of blowing or exhausting or both. For still lower pressures,
neasurable by water or mercury columns, fans are used of the disk or propeller
73
74 ENGINEERING THERMODYNAMICS
or centrifugal type. These fans are most used for ventilation of buildings an
mines, but a modification, based on the principles of the steam turbine reversec
and termed turbocompressors, is being rapidly adapted to such higher pres
sures as have heretofore required piston compressors.
When highpressure air is required for driving rock drills in mines and fc
hoisting engines, for tools, as metal drills, riveters, chipping chisels, for ca
air brakes, the compressors used to provide the air are termed simply ai
compressors. These compressors usually take in atmospheric air and compres
it to the desired pressure, the capacity required being usually adjustable
they have valves of the automatic type throughout commonly, but in larg
sizes frequently are fitted with mechanically operated suction valves t
decrease the resistance to entrance of air and so increase economy, a com
plication not warranted in small machines. When the pressures of deliver
are quite high the compression is done in stages in successive cylinders, th
discharge from the first or lowpressure cylinder being delivered through
water cooler or inter cooler to the second cylinder and occasionally to a thir
in turn. This staging with intermediate or intercooling results in bettc
economy, as will be seen later in detail, and permits the attainment of th
desired quantity of cool compressed air for subsequent use with the expenditur
of less work, the extra complication and cost being warranted only whe
machines are large and final pressures high.
In the operation of large steam condensers, noncondensible gases wi
collect and spoil the vacuum, which can be maintained only by the continuoi
removal of these gases, consisting of air, carbon dioxide and gases of anim;
and vegetable decomposition originally present in the water. When these gas<
are separately removed the machine used is a special form of compressor terme
a dry vacuum pump which, therefore, receives a charge at the absolute pre
sure corresponding to the vacuum, or as nearly so as the entrance resistano
permits, and after compression discharges into the atmosphere at a pressui
in the cylinder above atmosphere equivalent to discharge resistance. Natura
gas wells near exhaustion can sometimes be made to flow freely by the applic.
tion of a compressor capable of drawing a charge at a pressure below atmospher;
but whether the charge be received below atmospheric pressure or above i
in normal wells, the compressor will permit the delivery of the gas to distal
cities or points of consumption even 250 miles away through smaller pip
than would be otherwise possible. Naturalgas compressors, some steam ar
some gasengine driven, are in use for both these purposes, compressing natur
gas from whatever pressure may exist at the well to whatever is desired at tl
beginning of the pipe line.
In the preparation of liquid ammonia or carbonic acid gas for the marke
as such, or in the operation of refrigerating machinery, wet or dry vapor is con
pressed into a condenser to permit liquefaction by the combined effect of hij
pressure and cooling. One form of refrigerating machine merely compress
air, subsequently expanding it after preliminary cooling by water, so th;
after expansion is complete it will become extremely cold.
WOEK OF COMPRESSORS 75
All these compressing machines have, as a primary purpose, either the
>moval of a quantity of lowpressure gases from a given place, or the delivery
f a quantity of higherpressure gas to another place or both, but all include
Dmpression as an intermediate step between constantpressure admission and
onst antpressure discharge as nearly as structure may permit. They will all
ivolve the same sort of physical operations and can be analyzed by the same
rinciples except the wetvapor or wetgas compressors, in which condensation
<r evaporation may complicate the process and introduce elements that can
Jje treated only by thermal analysis later. Safe compressors cannot be built
rdth zero cylinder clearance, hence at the end of delivery there will remain in
ifhe clearance space a volume of highpressure gases equal to the volume of the
(Clearance space. On the return stroke this clearance volume will expand until
he pressure is low enough to permit suction, so that the new charge cannot
nter the cylinder until some portion of the stroke has been covered to permit
''his reexpansion of clearance gases.
It is quite impossible to study here all the effects or influences of structure
Ita indicated by the compressor indicator cards, but a quite satisfactory treat
nent can be given by the establishment of reference diagrams as standards of
Comparison and noting the nature of the differences between the actual cases and
'he standard reference diagram. These standard reference diagrams will really
>e pressurevolume diagrams, the phases of which correspond to certain hypoth
*!ses capable of mathematical expression, such as constant pressure, constant
Volume, expansion, and compression, according to some law, or with some
definite value of s fixing either the heatexchange character of the process or
fihe substance, as already explained.
2. Standard Reference Diagrams or PV Cycles for Compressors and Methods
)f Analysis of Compressor Work and Capacity. All the standard reference
diagrams will include constantpressure lines corresponding to delivery and
'supply at pressures assumed equal to whatever exists outside the. v cylinder on
lather delivery or suction side, that is, assuming no loss of pressure on delivery
br suction. The compression may be single or multistage with various
jimounts of cooling in the intercooler, but in multistage compression
ohe standard reference diagram will be assumed to involve intermediate
jooling of the gases to their original temperature, so that the gases
3ntering all cylinders will be assumed to have the same temperature and
X) maintain it constant during admission. Another difference entering into
Ijhe classification of standard reference diagrams is the laws of compression
is defined by the exponent s. Integration of the differential work expres
sion will take a logarithmic form for s = l, and an exponential form for all
Uther values, thus giving two possible reference compression curves and two
sets of work equations.
(a) The isothermal f or which s = 1 , no matter what the gas, and which is
the consequence of assuming that all the heat liberated by compression is con
tinuously carried away as fast as set free, so that the temperature cannot rise
it all.
76 ENGINEERING THERMODYNAMICS
(6) The exponential for which s has a value greater than one, general!
different for every gas, vapor or gasvapor mixture, but constant for any on
gas, and also for dry vapors that remain dry for the whole process. Wet vapor
having variable values of s cannot be treated by the simple pressurevolum
analysis that suffices for the gases, but must be analyzed thermally. Th
adiabatic value of s is a consequence of assuming no heat exchange at a
between the gas and anything else and is a special case of the general exponer
tial class.
Just why these two assumptions of thermal condition should result in th
specified values of s will be taken up under the thermal analysis part of thi
work, and is of no interest at this time.
As a consequence of these phase possibilities there may be established eigh
standard reference diagrams or pressurevolume cycles defined by their phase*
as shown in Fig. 23, four for singlestage compression and two each for two an
three stages. These might be extended by adding two more for four stage
and so on, but as it seldom is desirable, all things being considered, to go beyon
three, the analysis will stop with the eight cycles or reference diagrams showr
SINGLESTAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS
Cycle 1. Singlestage Isothermal Compression without Clearance.
Phase (a) Constant pressure supply.
" (6) Isothermal compression.
'" (c) Constant pressure delivery.
" (d) Constant zerovolume pressure drop.
Cycle 2. Singlestage Isothermal Compression with Clearance.
Phase (a) Constant pressure supply.
" (6) Isothermal compression.
" (c) Constant pressure delivery.
" (d) Isothermal reexpansion.
Cycle 3. Singlestage Exponential Compression without Clearance.
Phase (a) Constant pressure supply.
" (6) Exponential compression.
(c) Constant pressure delivery.
" (d) Constant zerovolume pressure drop.
Cycle 4. Singlestage Exponential Compression with Clearance.
Phase (a) Constant pressure supply.
(b) Exponential compression.
" (c) Constant pressure delivery.
" (d) Exponential reexpansion.
WORK OF COMPRESSORS
77
123456
Volumesja Cubic Feet
56 123456
aic Feet Volumes ia Cubic Feet
rs; One, Two, and ThreeStage, with and without Clearance,
onential Laws.
y
7
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r
3
Z
^ :; 
X
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j bg jad epunoj ui saanssaij
345
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glumes in Cubic Feet
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78 ENGINEERING THERMODYNAMICS
MULTISTAGE COMPRESSION
The phases making up multistage compression cycles may be consider
two ways, first, as referred to each cylinder and intercooler separately, b
second, as referred to the pressure volume changes of the gases themselve
regardless of whether the changes take place in cylinders or intercoolers.
For example, if 10 cu.ft. of hot compressed air be delivered from the firs
cylinder of 50 cu.ft. displacement, the phase referred to this cylinder is a con
stantpressure decreasing volume, delivery line whose length is ^ of the whoL
diagram, exactly as in singlestage compression. If this 10 cu.ft. of air delivj
ered to an intercooler became 8 ft. at the same constant pressure as the firs
cylinder delivery, the phase would be indicated by a constantpressure volum
reduction line 2 cu.ft long to scale, or referred to the original volume of ai
admitted to the first cylinder, a line gV of its length. Finally, admitting thi
8 cu.ft of cool air to the second cylinder and compressing it to  of its vo
ume would result in a final delivery line at constant pressure of a length of i cj
the length of the second cylinder diagram, but as this represents only 8 cu.ft
the final delivery will represent only X8 = 1.6 cu.ft. This 1.6 cu.ft. wilj
when referred to the original 50 cu.ft. admitted to the first cylinder, be repr<
1,6
sented by a constantpressure line, "rr = .032, of the whole diagram lengtl
oU
which in volume is equivalent to J of the length of the second cylind<:
diagram. It should be noted also that three volume change operations tali
place at the intermediate pressure; first, first cylinder delivery; secon<
volume decrease due to intercooling; third, second cylinder admission, tl
net effect of which referred to actual gas volumes, regardless of place whe
the changes happen, is represented by the volume decrease due to inte
cooling only; A diagram of volumes and pressures representing the resultai
of all the gas processes is called in practice the combined PV diagram for tl
two cylinders, or when plotted from actual indicator cards with due regai
for the different clearances of each cylinder the combined indicator diagran
It is proper in the study of the whole process of compression to consider tl
cycle consisting of phases referred to true gas volumes rather than phas,
referring to separate cylinder processes, which is equivalent to imagining tlj
whole cycle carried out in one cylinder.
Intercooling effects measured by the amount of decrease of volume j
constant pressure will, of course, depend on the amount of cooling or redu
tion of temperature, but in establishing a standard reference diagram sorj
definite amount capable of algebraic description must be assumed as .
intercooling hypothesis.
It has already been shown, Fig. 6, Chapter I, that from any original stf *
of pressure and volume the exponential and isothermal could be drawn, divergi fc
an amount depending on the difference between the defining exponent, L
If, after reaching a given state on the exponential curve, the gas be cooled J
WORK OF COMPRESSORS 79
mstant pressure to its original temperature, the point indicating its condition
ill lie by definition on the other curve or isothermal and the cooling process
represented by a horizontal joining the two curves. Such intercooling as
lis will be defined as perfect intercooling, for want of a better name, and its
ressurevolume effects can be treated by the curve intersections. It is now
ossible to set down the phases for the standard reference diagrams of multi
age compression, if in addition to the above it be admitted, as will be proved
,ter, that there is a best or most economical receiver pressure.
TWOSTAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS
Cycle 5. Twostage Exponential Compression without Clearance, Perfect
Intercooling at Best Receiver Pressure.
Phase (a) Constant pressure supply.
' ' (b) Exponential compression to best receiver pressure.
' ' (c) Constant pressure perfect intercooling of delivered gas.
' ' (d) Exponential compression from best receiver pressure.
' ' (e) Constant pressure delivery.
' ' (/) Constant zerovolume pressure drop.
Cycle 6. Twostage Exponential Compression with Clearance, Perfect
Intercooling at Best Receiver Pressure.
Phase (a) Constant pressure supply.
' ' (6) Exponential compression to best receiver pressure.
1 ' (c) Constant pressure perfect intercooling of delivered gas.
' ' (d) Exponential reexpansion of first stage clearance.
' ' (e) Exponential compression from best receiver pressure.
' ' (/) Constant pressure delivery.
' ' (g) Exponential reexpansion of second stage clearance.
THREESTAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS
Cycle 7. Threestage Exponential Compression, without Clearance, Per
fect Intercooling at Best Two Receiver Pressures.
Phase (a) Constant pressure supply.
1 ' (b) Exponential compression to first receiver pressure.
(c) Perfect intercooling at best first receiver presssure,
(d) Exponential compression from best first to best second
receiver pressure.
' ' (e) Perfect intercooling at best second receiver pressure.
' ' (/) Exponential compression from best second receiver pressure.
' ' (g) Constant pressure delivery.
" (h) Constant zerovolume pressure drop.
80 ENGINEERING THERMODYNAMICS
Cycle 8. Threestage Adiabatic Compression with Clearance, Perfec
Inter cooling at Best Two Receiver Pressures.
Phase (a) Constant pressure admission.
' ' (6) Exponential compression to best first receiver pressure.
' ' (c) Perfect cooling of delivered gas at best first receiver pressi
' ' (d) Exponential reexpansion of first stage clearance.
" (e) Exponential compression from best first to best sec<
receiver pressure.
1 ' (/) Perfect intercooling of delivered gas at best second recer
pressure.
' l (g) Exponential reexpansion of second stage clearance.
' ' (h) Exponential compression from best second receiver pressi
' l (i) Constant pressure delivery.
' ' (f) Exponential reexpansion of third stage clearance.
It should be noted that cycles 6 and 8 may be subdivided into any numbe
of cases, of which some of the most characteristic are shown: (a) where
.clearance volume in each cylinder bears the same ratio to the displacerru
of that cylinder, and commonly called equal clearances; (6) where the clearai
are. such that the volume after reexpansion in the higherpressure cylin<
is equal to the volume of clearance in the next lowerpressure cylinder, caus
the combined diagram to have a continuous reexpansion line, a case wl
may be called proportionate clearance; and (c) the general case in which tl
is no particular relation between clearances in the several cylinders.
By means of these definitions or their mathematical equivalents in syml
it will be possible to calculate work as a function of pressures and volui
and by various transformations of a general expresssion for work of a referei
cycle to calculate the horsepower corresponding to the removal of a gr
volume of gas per minute from the lowpressure supply or to the delivery
another volume per minute to the highpressure receiver or per unit weij
per minute. It will also be possible to calculate the necessary cylinder si;
or displacement per unit of gas handled, and the horsepower necessary 1
drive the compressing piston at a specified rate and further to calculate tl!
work and horsepower of cylinders of given size and speed. In order that thei
calculations of a numerical sort may be quickly made, which is quite necessaij
if they are to be useful, the formulas must be definite and of proper form, tl
form being considered proper when little or no algebraic transformation
necessary before numerical work is possible. While special expressions f<!
each case are necessary to facilitate numerical work, it is equally important
if not more so, to make clear the broad general principles or methods of attac.
because it is quite impossible to set down every case or even to conceive at tlj
time of writing of all different cases that must in future arise. The treaj
ment, then, must be a combination of general and special, the general metho I
being applied successively, to make them clear and as a matter of drill, nj
to every possible case, but only to certain characteristic or type forri
WORK OF COMPRESSORS 81
,f cases, such as are here set down as standard reference diagrams.
ndividual cases may be judged by comparison with these and certain factors
f relation established which, being ratios, may be and are called efficiencies.
.1ms, if a singlestage compressor should require two horsepower per cubic
oot of tree air compressed per minute, and Cycle I should for the same
>ressure limits require only one horsepower for its execution, then the efficiency
>f the real compression would be 50 per cent referred to Cycle I, and similar
actors or efficiencies for other compressors similarly obtained; a com
parison of the factors will yield information for a judgment of the two
:ompressors.
In what follows on the work and gas capacity of compressors two methods
j)f attack will be used.
1. General pressure volume analysis in terms of gas pressures and volumes
esulting in the evaluation of work per cubic foot of low or high pressure
i;aseous substance.
2. Transformation of results of (1) to yield volumetric efficiencies, mean
.effective pressures, work, horsepower, and capacity in terms of dimensions
of cylinders and clearances.
3. Singlestage Compressor, No Clearance, Isothermal Compression
[Cycle 1), Work, Capacity, and Work per Cubic Foot in Terms of Pressures
ind Volumes. The standard reference diagram is represented by Fig. 24, on
tfhich the process (A to B) represents admission or supply at constant pressure;
'B to C) compression at constant temperature; (C to D) delivery at constant
pressure; and (D to A) 2erovolume.
Let F& = The number of cubic feet of low pressure gas in the cylinder after
admission, represented to scale on the diagram by AB and equal
to the volume at B\
V c = volume in cubic feet of the gas in cylinder when discharge begins,
represented by DC, which is the volume at C;
" Pb = absolute pressure in pounds per square foot, at which supply
enters cylinder = (Sup. Pr.) = pressure at B;
pb = Pb+ 144 = absolute supply pressure in pounds per square inch =
(sup.pr.);
PC absolute pressure in pounds per square foot, at which delivery
occurs = (Del. Pr.) = pressure at C;
PC = PC 5 144 = absolute delivery pressure in pounds per square inch
= (del.pr.);
p
R p = = ratio of delivery pressure to supply pressure; v,
^
W = footpounds work done for the cycle; \
(H. P. Cap.) = volume of gas delivered in cubic feet per cycle, at
temperature same as that of supply;
(L. P. Cap. )= volume of gas drawn into cylinder, cubic feet per cycle.
For this no clearance case (L. P. Cap.) = F&.
82
ENGINEERING THERMODYNAMICS
Referring to Fig. 24, the work for the cycle is the sum of compression anc
delivery work, less admission work, or by areas
Net work A BCD = compression work EBCG+ deli very work GCDF
admission work EBAF
Algebraically this is equivalent to
1 A
FIG. 24. Onestage Compressor Cycle 1, No Clearance, Isothermal.
But since P c V c = PbVb the expression becomes
which is the work for the execution of the cycle when pressures and volui
are in pounds per square foot, and cubic feet. The equivalent expressh
for pounds per square inch and cubic feet is
W =144
Pb
WORK OF COMPRESSORS 83
Since, when there is no clearance the volume taken into the cylinder for
ach cycle (L. P. Cap.) is equal to the volume at B, Vb, the expression Eq. (29)
aay be stated thus, symbolic form.
TF=144(sup.pr.)(L. P. Cap.) loge R P (30)
The work per cubic foot of low pressure gas, footpounds, will be the above
xpression divided by (L. P. Cap.), or
I W
144 (sup.pr.) log, R p (31)
(L. P. Cap.)
The work per cubic foot of highpressure gas delivered will be
(H P ^~~ ^ = "^ ( su P*P r ') Rp ^& e RP>
mce
>r
vhich expressed symbolically is
.) = (H.P.Cap.)X#p. . ..... (33)
Expressions (31) and (32) for the isothermal compressor are especially useful
is standards of comparison for the economy of the compressors using methods
3ther than isothermal. It will be found that the work per cubic foot of either
ow pressure or cooled highpressure gas is less by the isothermal process than
3y any other process discussed later, and that it is the limiting case for the
economy of multistage compressors with a great number of stages. The fact
that this process of isothermal compression is seldom if ever approached in
practice does not make it any the less a suitable basis for comparison.
Example 1. Method of calculating Diagram Fig. 24.
Assumed Data.
p a =p b =, 2116 Ibs. per square foot. V a = V d =0.
P c =Pd = 18,000 Ibs. per square foot. Capacity =5 cu.ft.
To obtain point C,
Pb 5X2116
PC b ~ 18,000
.59, PC = 18,000.
84 ENGINEERING THERMODYNAMICS
Intermediate points B to C are obtained by assuming various pressures and findir
corresponding volumes as for V c .
Example 2. To compress and deliver 5 cu.ft. of air from atmospheric pressure (211
Ibs. per square foot) to 8.5 atmospheres (18,000 Ibs. per square foot) isothermal]
without clearance, how much work is necessary?
P 6 =2116 P c = 18,000
F 6 =5
 8 ^ 17 CiSft
Vc~Pb d
Work of admission =P b Vi> =2116 X5 = 10,585 ft.lbs.
p
Work of compression =P b V b log e ~ =10,585 Xlog e 8.5 =22,600 ft.lbs.
f b
Work of delivery =P C V C = 10,585 ft.lbs.
Total work = 10,585 + 22,600  10,585 = 22,600 ft.lbs.
Or by the general formula,
TF = (sup.pr.)(L.P.Cap.) log e R p =2116x5 xlog e 8.5 =2116x5x2.14 =22,652 ft.lb,
Prob. 1. How many cubic feet of free air may be compressed and delivered pc
minute from 14 Ibs. absolute to 80 Ibs. per square inc 11 , absolute per horsepower in
compressor with zero clearance if compression is isothermal?
Prob. 2. Gas is being forced through mains at the rate of 10,000 cu.ft. per minul
under a pressure of 5 Ibs. per square inch above atmosphere. The gas is taken into tt
compressor at atmospheric pressure and compression is isothermal. What hors<
power will be needed at sea level and at an elevation of 5000 feet?
Prob. 3. Natural gas is drawn from a well, compressed isothermally and forced throug
a main at the rate of 200,000 cu.ft. per hour measured at the pressure on the suctio
side. What steam horsepower will be required to operate the compressor if th
mechanical efficiency be 80 per cent? Suction pressure is 8 Ibs. per square inc
absolute, delivery pressure 60 Ibs. per square inch absolute.
Prob. 4. A vacuum cleaning pump is required to maintain a pressure of 14 Ibs. p(
square inch absolute, move 500 cu.ft. of free air per minute and discharge it againt
an atmospheric pressure of 15 Ibs. per square inch absolute. What horsepower wi
be required (isothermal)?
Prob. 5. A blower furnishes 45 cu.ft. of ir a minute at a pressure of 5 ins. of mercur
above atmosphere. Assuming compression to be isothermal and supply pressure to t
atmospheric, what horsepower will be needed?
Prob. 6. A compressor has a piston displacement of 3 cu.ft. At what speed can it I
run if air be compressed isothermally from 1 to 10 atmospheres and the horsepow<
supplied is 100?
Prob. 7. A tank of 1000 cu.ft. capacity contains air at atmospheric pressure. .
compressor taking air from atmosphere compresses it isothermally and discharges it inl
the tank until the pressure reaches 100 Ibs. per square inch gage. What horsepow(
will be required to fill tank at this pressure in ten minutes?
Prob. 8. A compressor receives air at atmosphere and compresses it isothermally i
five atmospheres above atmosphere. It takes in 1000 cu.ft. of free air per minut*
How much would the capacity increase if the discharge pressure dropped to 3 atmo
pheres and the horsepower remained the. same?
WORK OF COMPRESSORS
85
Prob. 9. Suppose that the pressure in the above problem were raised to 8 atmos
Ueres. How much would the capacity decrease if the horsepower remained the same
id how much more power would be required to keep the capacity the same?
Prob. 10. By means of suitable apparatus, the water from the side of a waterfall is
verted to a vertical shaft, and in falling 126 ft. compresses air from atmospheric pressure
a value equal to 90 per cent of the head of the water. To deliver 1000 cu.ft. of com
essed air per hour, how much water is required if the work of falling water is 80 per
;;nt useful in compressing the air?
4. Singlestage Compressor with Clearance, Isothermal Compression,
2ycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures
ad Volumes.
FIG. 25. OneStage Compressor Cycle 2, Clearance, Isothermal.
Referring to Fig. 25, the work of the cycle i^by areas.
Net work
HADFEBAH
= Area ABC D. .
[t is easily seen that this area is also equal to (JBCL) (JADL), both
which are areas of the form evaluated in the preceding section Accordingly
Net work
 JADL,
86 ENGINEERING THERMODYNAMICS
Algebraically,
(3
which is the general expression for the work of the cycle in footpounds whi
pressures are in pounds per square foot, and volumes in cubic feet. Substituti]
the symbolic equivalents and using pressures in pounds per square inch, the
results, since (F& F a ) = (L. P. Cap.),
Work = 144 (sup.pr.)(L. P. Cap.) log e R p , .... (3
which is identical with Eq. (30), showing that for a given lowpressure capaci
the work of isothermal compressors is independent of clearance. The value
the lowpressure capacity (F& V a ) may not be known directly, but m;
be found if the volume before compression, F&, the clearance volume befc
reexpansion, Vd, and the ratio of delivery to supply pressure, R p , are know
thus
V a =V d R p)
from which
(L. P. Cap.) = (VV Ftffl,) ......... (2
From Eq. (35) the work per cubic foot of lowpressure gas is, in footpounds,
W
144(sup.pr.)
(L.P. Cap.)
and the work per cubic foot of highpressure gas delivered, ft.lbs.
W
(H.P. Cap.)
144(sup.pr.)# p log e R
By comparison, Eqs. (37) and (38) are found to be identical with (31) a
(32) respectively, since clearance has, as found above, no effect on the work del
for a given volume of gas admitted, however much it may affect the work of i
cycle between given volume limits or work per unit of displacement.
It is interesting to note that the work areas of Figs. 24 and 25 are eqi
when plotted on equal admission lines AB or delivery lines CD and aj
horizontal intercept xy will be equal in length on both if drawn at thesa:f
pressure.
In what precedes, it has been assumed that AB represents admission voluii
and CD represents delivery volume which is true for these established eye I
WORK OF COMPRESSORS 87
sference, but it is well to repeat that for real compressors these are only
>parent admission and delivery lines, as both neglect heating and cooling
.fects on the gas during its passage into and out of the cylinder. Also that
: real compressors the pressure of the admission line cannot ever be as high
: T, the pressure from which the charge is drawn and the delivery pressure must
jj necessarily higher than that which receives the discharge, in which cases
,ie volume of gas admitted, as represented by AB, even if the temperature
id not change, would not equal the volume taken from the external
ijipply, because it would exist in the cylinder at a lower pressure than it
,'iginally had, and a similar statement would be true for delivered gas.
! Problems. Repeat all the problems of the last section, assuming any numerical
jilue for the clearance up to 10 per cent of the displacement.
i 5. Singlestage Compressor Isothermal Compression. Capacity, Volu
metric Efficiency, Work, Mean Effective Pressure, Horsepower and Horse
ower per Cubic Foot of Substance, in Terms of Dimensions of Cylinder and
learance.
Consider first the case where clearance is not zero. Then Fig. 25 is the
;jference diagram.
Let D = displacement = volume, in cubic feet, displaced by piston in one
stroke = area of piston in sq.ft. X stroke in ft. = (F& V d ).
" (H. P. Cap.) = high pressure capacity = vol. cu.ft. of gas delivered per
cycle at temperature equal to that of supply =(F C V d );
(L. P. Cap.) = low pressure capacity = vol. in cu.ft. of gas entering
cylinder per cycle =(F& F a );
L.P. Cap. V b V a
E v = volumetric efficiency = ^j = y _y >
" Cl = volume of clearance, cubic feet= V d
" c = clearance volume expressed as a fraction of the displacement;
Cl V
=~. = v * whence Cl = cD',
LJ V b v a
W
M.E.P. =mean effective pressure, Ibs. per square foot = yr;
W
m.e.p. = mean effective pressure, Ibs. per square inch = ~ ;
" N number of revolutions per minute;
" n number of cycles per minute;
N
z = number of revolutions per cycle = ;
" I.H.P. = indicated horsepower of compressor;
88 ENGINEERING THERMODYNAMICS
The lowpressure capacity of the singlestage isothermal compressor wit
clearance is,
but
Whence (L. P. Cap.) = ( V b V d j/ j for which may be substituted the symbo
for displacement and clearance volumes, thus
(L. P. Csip.)=D+cDcDR p ,
= D(\+ccR p )
For convenience the term, Volumetric Efficiency, E v is introduced. Sim
this is defined as the ratio of the lowpressure capacity to the displacement,
(K P. <*>.)_,
Referring to Eq. (35) it is seen that the value of (L. P. Cap.) can 1
substituted from Eq. (39) and the result is:
Work per cycle, footpounds, in terms of supply pressure, pound spersq ua
inch, displacement cubic feet, clearance as a fraction of displacement, and rat
of delivery to supply pressure is,
HF = 144 (sup.pr.)D(l+ccRp)\ogeR
or in the terms of the same quantities omitting clearance and introduce
volumetric efficiency, E v ,
/ TF=144(sup.pr.) DE v \og e R P , . . (4
To obtain the mean effective pressure for the cycle, the work done per cy<
is divided by displacement, D.
Mean effective pressure, pounds per square inch,
W
(m.e.p.) =
144D'
whence
(m.e.'p.) = (sup.pr.)(l+c cR P ) \og e R P
or
(m.e.p.) = (sup.pr.)^,, log c R p * . ( z
WORK OF COMPRESSORS 89
The indicated horsepower of the isothermal compressor is equal to the
work per minute, in ft.lbs. divided by 33,000. If n cycles are performed
per minute, then
(45)
(46)
^Introducing the effective area of the piston, in square inches, a, and the piston
Jspeed S, feet per minute, then since
44 2z
(sup.pr> ,
LILR  66,000 7^' logejRp ' ' ...... >.:'
The same expression for the indicated horsepower may be derived by the
.substitution of the value of (m.e.p.) Eq. (44) in the following general expres
ision for indicated horsepower.
= (m.e.p.)o
33000 X 2z
Example 1. Method of calculating Diagram Fig. 25.
Assumed Data:
P a =Pb =2116 Ibs. per square foot;
P d =P c =18,000 Ibs. per square foot;
c =3 per cent. L.P. Cap. =5 cu.ft. s = 1.
To obtain point D:
From formula Eq. (39), L.P. Cap. =D(l+ccRp) or 5=Z)(l+.03.03x8.5),
or
D =6.5 cu.ft. and Cl. =.03 X6.5 =.195 or approximately .2 cu.ft.
/. V d = .2 cu.ft., P d = 18,000 Ibs. sq.ft.
To obtain point A :
PaV a =P d V d or 7 a =!^ = 8.5X.2 = 1.7,
* a
.'. F a = 1.7 cu.ft., P a =21161bs. sq.ft.
90
ENGINEERING THERMODYNAMICS
Intermediate points D to A are obtained by assuming various pressures and finding the
corresponding volumes as for F a .
To obtain point B:
=6.7 cu.ft.
= 2116 Ibs. sq.ft.
To obtain point C:
':. V c = .79 cu.ft.,
PC = 18,000 Ibs. sq.ft.
Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to
atmospheres isothermally in a compressor having 4 per cent clearance. What mi
be the displacement work, per 100 cu.ft. of supplied and delivered air, and horsepowe];
of machine? Speed is 150 R. P.M., compressor is double acting and stroke = 1.5 diameters
Neglect piston rods.
D = (L. P. Cap.) ;#, and # = (!+. 04 .04x8.5) =.7,
.'. D = 1000 4 .7 = 1428 cu.ft. per minute.
Work per cu.ft. of supplied air = (sup.pr.) 144 log e R p = 144 X 14.7 log e R P =4530 ft.lbs.;
Hence the work per 100 cu.ft. =453,000 ft.lbs.
Work per cu.ft. of delivered air = 144(sup.pr.)# p log e R p = 144 X14.7 X8.5 X2.14 =38,^
ft.lbs.
Hence the work per 100 cu.ft. =3,855,000 ft.lbs.
D =
33000
1428
150X2
= 4.76 cu.ft. per stroke.
Hence cylinder diameter =
= 1.59 feet = 19.1 inches.
Prob. 1. How many cubic feet of free air per minute may be compressed isotl
mally to 100 Ibs. per square inch absolute in a compressor having 6 per cent clean
if the horsepower supplied is 60?
Prob. 2. A compressor has a cylinder 18x24 ins., clearance 4 per cent, is doul
acting and runs at 150 R.P.M. If it compresses air from atmosphere to 100 Ibs.
square inch gage, what will be its high and lowpressure capacity, its horsepower,
WORK OF COMPRESSORS 91
fiow will the horsepower and the capacity compare with these quantities in a
V hypothetical compressor of the same size but having zero clearance? How will
;he horsepower per cubic foot of delivered air compare?
Prob. 3. A manufacturer gives for a 10xl2 in. doubleacting compressor running
it 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50100 Ibs.
3er square inch gage. What clearance does this assume for the lowest and highest
pressure if the compression is isothermal? The horsepower is given as from 23 to 35.
r'dheck this.
Prob. 4. Air enters a compressor cylinder at 5 Ibs. per square inch absolute and is
compressed to atmosphere (barometer = 30 ins.). Another compressor receives air
rat atmosphere and compresses it to 3 atmospheres. If each has 6 per cent clearance
,:what must be the size of each to compress 1000 cu.ft. of free air per minute, how will
ijthe total work compare in each machine, and how will the work per cubic foot of high
;jand low pressure air compare in each? Assume compression to be isothermal.
Prob. 5. 1800 cu.ft. of free air per minute is to be compressed isothermally to apres
jsure of 100 Ibs. per square inch gage. What must be the displacement and horsepower
of a hypothetical zero clearance compressor, and how will they compare with those of a
i compressor with 6 per cent clearance?
Prob. 6. Consider a case of a compressor compressing air isothermally from atmos
phere to 100 Ibs. per square inch gage. Plot curves showing how displacement and
horsepower will vary with clearance for a 1000 cu.ft. free air per minute capacity
taking clearances from 1 per cent to 10 per cent.
Prob. 7. Two compressors of the same displacement^ namely 1000 cu.ft. per min
ute, compress air isothermally from 50 Ibs. per square inch gage to 150 Ibs. per square
inch gage. One has 5 per cent clearance, the other 10 per cent. How will their capaci
ties and horsepower compare with each other and with a no clearance compressor?
Prob. 8. A 9 Xl2 in. compressor is compressing air from atmosphere to 50 Ibs. gage.
How much free air will it draw in per stroke, and how much compressed air will it dis
charge per stroke for each per cent clearance?
Prob. 9. The volumetric efficiency of a compressor is .95 as found from the indicator
card. It is double acting and has a cylinder 18 X 24 ins. What will be its capacity and
required horsepower for 100 Ibs. per square inch gage delivery pressure?
6. Singlestage Compressor, No Clearance Exponential Compression,
(Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures
and Volumes. The cycle of the singlestage exponential compressor without
clearance is represented by Fig. 26. Referring to work areas on this diagram,
Net work A BCD = compression work EBCG
+ deli very work GCDF
admission work FARE.
Algebraically,
s 1
92
But
or
and
ENGINEERING THERMODYNAMICS
P V T7 *l p V V s 1
* ' c V c * J ' 6 ' & >
l]
18000
14400
(H.P. Cap 1
* Cold) *'(H.P. Cap,
Hot) j
, 
D

I
m ;00
^(Sup.Pr^
i
1
!
:
\
1
f
\
1
3
\
\
\
\
\
Pressure? in Pounds per Square
\
\

&
V
X
\
5
^
s
\"<
^
\
(S
k V.
$
\
&
Sa
A^
\
X< <X
^
\
'?
s
\
\
\
w
s
\
\
V
^
X
s
s
^
.^
*s.

^,
"V
s.
*^,
^
^^
^.
^
**.
^^
^:
s^
s
=H=
1 ? 3 4 5
* Volumes in Cubic Feet
1
L.P.
FIG. 26. OneStage Compressor Cycle 3, No Clearance, Exponential.
Substituting above
WORK OF COMPRESSORS 93
Thence
<**>
iCq. (48) gives the work in footpounds for the execution of the cycle when
)ressures are in pounds per square foot, and volumes in cubic feet.
The equivalent expression for pressures in pounds per square inch is
s 1
(49)
When there is no clearance, as before, F& represents the entire vol
}ime of displacement, which is also here equal to the volume admitted
.
!L. P. Cap.), p b is the supply pressure (sup.pr.) pounds per square inch
iibsolute and is the ratio of delivery to supply pressure, R p .
Accordingly, the work of an exponential, singlestage compressor with no
Clearance is
sl
Tf = 144^ T (sup.pr.)(L.P.Cap.)(^/ l) (50)
S L \ /
The work per cubic feet of low pressure gas, footpounds is
81
W s ( ~T~ \
/T p ^ v 144 (sup.pr.) I R P 1 (51)
(L. P. Cap.) s 1 \
n
Before obtaining the work per cubic foot of highpressure gas, it is neces
ary to describe two conditions that may exist. Since the exponential com
jiression is not isothermal, it may be concluded that a change in temperature
fvill take place during compression. This change is a rise in temperature,
jind its law of variation will be presented in another chapter.
1. If the compressed air is to be used immediately, before cooling takes
i)lace, the highpressure capacity or capacity of delivery will be equal to the
volume at C, V c and may be represented by (H. P. Cap. hot).
2. It more commonly occurs that the gas passes to a constantpressure
lolder or reservoir, in which it stands long enough for it to cool approximately
o the original temperature before compression, and the volume available
liter this cooling takes place is less than the actual volume discharged from
he cylinder in the heated condition. Let this volume of discharge when
educed to the initial temperature be represented by (H. P. Cap. cold) which
s represented by Vt, Fig. 26.
94 ENGINEERING THERMODYNAMICS
Since B and C in Fig. 26 lie on the exponential compression line, P&TV
or
(L. P. Cap.) = (H. P. Cap. hot) (R,)? (52
Hence, the work in footpounds per cubic foot of hot gas delivered fron 1
compressor is
W s i / si
/~r~r pj ^r~^ V~ TV *""~ * i A T yoLlJ3JJi * J f\jp ^ \ **'' p ^
(ti. r. uap. notj s i \
On the other hand, B and K lie on an isothermal and
since P* = P c ,
whence
(L. P. Cap.) = (H. P. Cap. cold)# p (54
The work footpounds per cubic foot of gas cooled to its original temper*
ture is, therefore,
W s / tzl \
= 144 (sup.pr.)R p (R p ij, . . . . (5f
(H. P. Cap. cold) sl
or
W
. . . . (5H
This last equation is useful in determining the work required for the storir
or supplying of a given amount of cool compressed air or gas, under conditioi
quite comparable with those of common practice.
Example 1. Method of calculating Diagram Fig. 26.
Assumed Data:
p a =p b =2116 Ibs. per square foot; P c =Pa= 18,000 Ibs. per square foot.
Cl =0; V a = V d =0; L. P. Capacity =5 cu.ft.; s = 1.4 (adiabatic value of ; J
To obtain point C:
or
_
P 6 /P & =8.5; Iog e 8.5 = .929, and .71 log* 8.5 = .665; (P c /P 6 )i4 =4.6,
WORK OF COMPRESSORS 95
lence V c = 5 J4.6 = 1.09 cu. ft. P c = 18,000 Ibs. per sq.ft.
.ntermediate points B to C are obtained by assuming various pressures and finding
,he corresponding volumes as for V c .
Example 2. To compress 5 cu.ft.of air from atmospheric pressure (2116 Ibs. per
quare foot) to 8.5 atmospheres (18,000 Ibs. per square foot) adiabatically and with no
learance requires how many footpounds of work?
P b =2116 Ibs. sq.ft.,, P c = 18,000 Ibs. sq.ft.,
F 6 =5 cu.ft.
F C =T]JT n =54.57 = 1.092 cu.ft.
^ork of admission is
P b V =2116X5 = 10,585 ft.lbs.
Work of compression, using y to represent the adiabatic value of s is,
Work of delivery is
P C V C = 18,000 X 1.092 = 19,650 ft.lbs.
:otal work = 19,650+22,35010,585=31,425 ft.lbs.,
>y the formula Eq. (50) directly
/ 1 \
TT = 144^j(sup. pr.) (L. P. Cap.) \R p r l.j
= 144+3.46 X2116 X5 X[(8.5)' 29 l];
= 144+3.46 X2116 X5 X.86 =31,450 ft.lbs.
rob. 1. A singlestage zero clearance compressor compresses air adiabatically from
6 atmospheres. How many cubic feet of free air per minute can be handled if the
iompressor is supplied with 25 H.P. net?
Prob. 2. The same compressor is used for superheated ammonia under the same
pressure conditions. For the same horsepower will the capacity be greater or less
ad how much?
Prob. 3. A dryvacuum pump receives air at 28 ins. of mercury vacuum and delivers
; against atmospheric pressure. What will be the work per cubic foot of lowpressure
ir and per cubic foot of highpressure air hot? Barometer reads 29.9 ins.
96 ENGINEERING THERMODYNAMICS
Prob. 4. The manufacturer gives for a 10iXl2 in. double acting compressor run
ning at 160 R.P.M., a capacity of 177 cu.ft. of free air per minute and a horsepower o
25 to 35 when delivering against pressures from 50 to 100 Ibs. Check these figures.
Prob. 5. A set of drills, hoists, etc., are operated on compressed air. For their opera
tion 3000 cu.ft. of air at 70 Ibs. gage pressure are required per minute. What must br
the piston displacement and horsepower of a compressor plant to supply this air i
compression is adiabatic and there is assumed to be no clearance?
Prob. 6. Air is compressed from atmosphere to 60 Ibs. per square inch gage by iji
compressor having a 12x18 in. cylinder and running at 100 R.P.M. Find its capacity
and horsepower at sea level and loss in capacity and horsepower if operated at an alti
tude of 10,000 ft. for zero clearance.
Prob. 7. 10,000 cu.ft. of free air per minute at a pressure of 15 Ibs. above atmos,i
phere are compressed and delivered by a blowing engine. Find the horsepower requires 1
to do this and find how much free air could be delivered by same horsepower if thj
pressure were tripled.
Prob. 8. In a gas engine the mixture of air and gas is compressed in the cylinde
before ignition. If the original pressure is 14 Ibs. per square inch absolute, fim
pressure 85 Ibs. absolute and compression is adiabatic, what will be the work c
compression only, per pound of mixture?
NOTE: Weight per cubic foot may be taken as .07 and y as 1.38.
Prob. 9. A vacuum pump is maintaining a 25in. vacuum and discharging the a
removed against atmospheric pressure. Compare the work per cubic foot of low pre:
sure air with that of a compressor compressing from atmosphere to 110 Ibs. above atmo:
phere.
7. Singlestage Compressor with Clearance, Exponential Compressioi
(Cycle 4). Work, Capacity, and Work per Cubic Foot in Terms of Pressure
and Volumes. When clearance exists in the cylinder, it is evident that
volume equal to the clearance, V*, will not be expelled during the deliver
of compressed gas, and this volume will expand with fall in pressure as tl
piston returns, causing pressurevolume changes represented by the line DJ
on the diagram, Fig. 27. Until the pressure has fallen to that of supply, tl!
admission valve will not open, so that while the total volume in the cylind<
at end of admission , is 7 6 , the volume V a was already present by reasoj
of the clearance, and the volume taken in is (F& Va) which is the lowpressui*
capacity (L. P. Cap.).
The work area of the diagram is A BCD, which may be expressed as
Work area = JBCL JADL,
which areas are of the form evaluated in Section 6. Hence, the above expressic
in algebraic terms is
WORK OF COMPRESSORS
97
is is the general expression for the work of the cycle, in footpounds, when
the pressures are expressed in pounds per square foot, and volumes in cubic
feet. Using symbolic equivalents
(H.P.'Cap.Hot)
1, .... (58)
18000
234
Volumes in Cubic Feet
U.P. Cap.)
D
FIG. 27. OneStage Compressor Cycle 4, Clearance, Exponential.
(58) is identical with Eq. (50), showing that for adiabatic as for
lermal compressors, the worl done for a given lowpressure capacity is inde
pendent of clearance. Due to this fact, the expressions derived for the expo
nential compressor without clearance will hold for that with clearance:
Work, in footpounds per cubic foot of lowpressure gas is,
W ' s
(L.P.Cap.r s ^i' '"
Work, in footpounds per cubic foot of hot gas delivered is,
(59)
(H.P.
81
s~
98 ENGINEERING THERMODYNAMICS
Work, in footpounds per cubic foot of cooled gas to its original temperature is,
= 144^(del.pr.)(W l] . (61)
(H. P. Cap. cold) sl
The relation of highpressure capacity either hot or cold to the lowpressure
capacity is also as given for the case of no clearance, as will be shown.
In Fig. 27, the highpressure capacity, hot, is DC=V c V d . The low
_ 11 11
pressure capacity is AB= V b V a , but VcP c * = V b Pb s and TW = TW, or
V b = VcR^ and V a = Va
i
Hence (L. P. Cap.) = (H. P. Cap. hot)tf/ ...... (62)
If the delivered gas be cooled to its original temperature, then the volume
after delivery and cooling will be
. (L. P. Cap.)
(H. P. Cap. cold) =  5  ,
Lip
or
(L.P. Cap.) = (H. P. Cap. cold)R p .... (63)
From the work relations given above, it is seen that in general, the work
per unit of gas, or the horsepower per unit of gas per minute is independent
of clearance.
8. Singlestage Compressor Exponential Compressor. Relation between
Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, Horse
power and H.P. per Cubic Foot of Substance and the Dimensions of Cylinder
and Clearance. As indicated on Fig. 27, for the singlestage exponential com
pressor with clearance, the cylinder displacement D, is (Vb Vd). The low
pressure capacity per cycle is (L. P. Cap.) = (F& F a ). The actual volume of
gas or vapor delivered by the compressor is (H. P. Cap. hot) = (V c ~ Vd).
This is, in the case of a gas at a higher temperature than during supply,
but if Qooled to the temperature which existed at B will become a less volume.
This delivered volume after cooling is symbolized by (H. P. Cap. cold) and is
equal to (L. P. Cap.) Xy r or ' = ' where R p is the ratio of delivery
(del.pr.) Hp
pressure to supply pressure.
Volumetric efficiency, E v , already defined as the ratio of lowpressure
capacity to displacement is
ny a _(L.P.Cap.)
*"V^V d  ~D '
Cl
WORK OF COMPRESSORS 99
earance, c, expressed as a fraction of the displacement is the ratio of
irance volume, Cl, to displacement, D, and is,
Cl__ V d
= D'V b V d
iean effective pressure, pounds per square foot (M.E.P.), is the mean height
Oithe diagram or the work area W, divided by displacement, D. If expressed
abounds per square inch the mean effective pressure will be indicated by
 Let (I.H.P.) be indicated horsepower of the compressor;
N the number of revolutions per minute;
n the number of cycles per minute and
z the number of revolutions per cycle, whence nXz =
I Then, the lowpressure capacity is
(L.P.
U
3ice the reexpansion DA is exponential and similar to compression as to
ilue of s, whence j
(L. P. Cap.) = (Vt,Va) = V b  V d R p ;
i
D+cDcDR P * ;
(L. P. Cap.)=D(l+cc3) (64)
.
*om this, by definition, the volumetric efficiency is
(L.P. Cap.) i f .
J% = 7: =l+c cR p * (65)
Referring to Eq.(57), in which may be substituted the value Eq. (64) for
7 & F a ), the work of the singlestage exponential compressor in terms of dis
100 ENGINEEEING THERMODYNAMICS
placement, clearance (as a fraction of displacement), and pressures of suj
and delivery in pounds per square foot is,
or using pressures, pounds per square inch, and inserting the symbols, thi
may be stated in either of the following forms:
w=
c/^) \Rp  l. . . .
The mean effective pressure in pounds per square foot is this work dividfj
by the displacement, in cubic feet, and may be converted to pounds per squa
inch by dividing by 144, whence
Mean effective pressure, pounds per square inch.
6
(7
The indicated horsepower of the singlestage exponential compressor frcjl
(67) is,
T H P  Wn . s (sup.pr.)nDE v [ ^_ 1 (
LKR 33000"^! ~ ~^2&r
Where n is the number of cycles per minute, or in terms of piston speed S a I
effective area of piston, square inches, and z the number of revolutions
cycle, i
s (sup.pr.)qff p I" i^ 1 1
[  66()()02 ^
Since it was found in Section 7, that the work per unit volume of gas is t
same with clearance as without clearance, the horsepower per cubic foot i
minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ).j
Horsepower per cubic foot of gas supplied per minute
IH.P. _!_fouprOU ^J
n(L.P.Cap.)~l 229:2 Kv
WOKK OF COMPRESSORS, ' 101
The horsepower per cubic foot of hot gas delivered per minute is
LH.P. (sup.pr.) I f =! 1
n(H.P.Cap.hot) = ^I 229.2 R "' * *
brsepower per cubic foot of gas delivered and cooled is
IH.P. _ __ (sup.pr.) T til _ I
n(H.P. Cap .cold) ~sl 229.2 ^[/^ ' l \' '
s (del.pr.Ho ,1
= . 122932 I*' 8 ' ' < 76 >
i In the above formulae (del.pr.) and (sup.pr.) indicate delivery pressure and
spply pressure, in pounds per square inch.
Example 1. Method of calculating Diagram, Fig. 27.
\3sumed data:
p a =p b =2116 Ibs. per square foot.
p c =p d = 18,000 Ibs. per square foot.
Cl. =3.5 per cent. L. P. Capacity =5 cu.ft. s = 1.4.
'3 obtain point D:
/ i \ / \ .715
L. P. Ca,p.=D\l+ccR7) or 5=D U +.035 .035 (8.5) / ;
ence
D5^(1+.035.035X4.6) =5.72 cu.ft. and C7 = . 035x5.72 =.2 cu.ft
.*. V d = .2 cu.f t. ; P d = 18,000 Ibs. sq.ft.
jo obtain point A :
= 4.6X.2=.92;
/. Fa =.92 cu.ft.; P a =2116 Ibs. sq.ft.
itermediate points D to A are obtained by assuming various pressures arid finding the
^responding volumes as for V a .
o obtain point B:
. P. Cap. =.92 +5 =5.92.
/. F & = 5.92 cu.ft.; P b =2116 Ibs. sq.ft.
ENGINEERING THERMODYNAMICS
To obtain point C:
= 5.92 =4.6 = 1.29 cu.ft.
.'. V c = 1 .29 cu.ft. ; PC = 18,000 Ibs. sq.ft.
Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to ':
atmospheres absolute so that s = 1.4, in a compressor having 4 per cent clearance. Wl t
must be the displacement of the compressor, work per 100 cu.ft. of supplied and deliver i
air, hot and cold, and horsepower of machine? Speed is 150 R.P.M., compressor
double acting and stroke = 1.5 diameters.
D =L. P. Cap. +E V , and E v = (l f c cR^ ) .
.*. # p =(l+.04.04x(8.5) 71 ) = .86;
/. D = 1000 + .86 = 1 162 cu.ft. per min.
.sl
Work per cubic foot of supplied air = 144   (sup.pr)[/tV s 1],
s 1
= 144 X3.46X 14.7 X. 86 =6300 ft.lbs.
.'. Work per 1000 cu.ft. = 6,300,000 ft.lbs.
Work per cubic foot of delivered air cold is R p times work per cubic foot of suppliec
hence work per 100 cu.ft. of delivered cooled air is 5,350,000 ft.lbs.
Work per cubic foot of delivered air hot is R p ~^ times work per cubic foot of supplied
hence work per 100 cu.ft. of hot delivered air is 2,800,000 ft.lbs.
or
(m.e.p.) = 3.46 X 14.7 X. 86 X. 86 =37.7 Ibs. per square inch.
.'. ^=5690 or d = 17.85.
a =250 sq.inches. S =670 ft. per min.
.'. I.H.P.=191.
WORK OF COMPRESSORS io<
Prob. 1. A denseair ice machine requires that 4000 cu.ft. of air at 50 Ibs. per squar<
inch absolute be compressed each minute to 150 Ibs. per square inch absolute. The
compression being such that s = 1.4, clearance being 6 per cent, find the work required
What would be the work if clearance were double? Half?
Prob. 2. The compressor for an ammonia machine compresses from one atmos
phere to 8 atmospheres absolute. With adiabatic compression and 4 per cent clear
ance, what will be work per cubic foot of vapor at the low pressure and at the high!
Assume vapor to be superheated.
Prob. 3. On a locomotive an airbrake pump compresses air adiabatically frocc
atmosphere to 80 Ibs. per square inch gage. It is required to compress 50 cu.ft. of free
air per minute and clearance is 5 per cent. What horsepower must be supplied to it!
Prob. 4. In a manufacturing process a tank must be maintained with a vacuum oi
29 ins. when barometer reads 30 ins. To do this 100 cu.ft. of carbon dioxide must be
removed from it per minute and returned under atmospheric pressure. Compression is
adiabatic and clearance 7 per cent. How much power must be supplied to compressor
and what should be its displacement?
Prob. 5. Two compressors each 12x18 in., double acting, with 8 per cent clear
ance, and running at 150 R.P.M. compress in the one case air, in the other carbon disul
phide. The compression being adiabatic in each case, what (a), is the difference in
power required, (6), in lowpressure capacities? Take pressures as 2 and 15 atmos
pheres of 26 inches mercury.
Prob. 6. A compressor is supplied with 40 horsepower. If it draws in air from
atmosphere to what pressure can if 500 cu.ft. per minute be compressed, when s = 1.38
and clearance 10 per cent?
Prob. 7. For forcing gas through a main, a pressure of 50 Ibs. per square inch gage
is required. W'hat is the work done per cubic foot of highpressure gas, if a compressor
having 6 per cent clearance is used, and s for the gas is 1.36? What should be its dis
placement?
Prob. 8. A gas compressor 20x22 ins. has a volumetric efficiency of 90 per cent,
supply pressure =4 Ibs. per square inch and delivery 110 Ibs. per square inch gage.
W 7 hat are its L. P., H. P., hot and cold capacities, and its I.H.P. if single acting at 70
R.P.M. when s = 1.35?
9. TwoStage Compressor, no Clearance, Perfect Intercooling, Exponential
Compression, Best Receiver Pressure, Equality of Stages (Cycle V). Work and
Capacity in Terms of Pressures and Volumes. The common assumption in con
sidering the multistage compressor is that in passing from one cylinder to the
next, the gas is cooled to the temperature it had before entering the com
pressor, which has already (Section 2), been defined as " perfect inter cooling."
This condition may be stated in other words by saying that the product
of pressure and volume must be the same for gas entering each cylinder. If
then the volume and pressure of gas entering the first stage be determined,
fixing the volume entering the second stage will determine the pressure of the
gas entering the second stage, or fixing the pressure of the gas entering the
second stage will determine the volume that must be taken in.
Using subscripts referring to Fig. 28, for the no clearance case,"
P b V, = P d V d (77)
104 ENGINEERING THERMODYNAMICS
The net work of the compressor, area ABCDEF = area ABCH first stage
area HDEF second stage. Using the general expression, Eq. (48) for these
work areas with appropriate changes in subscripts
81
'
stage)
Sl
+^^ p dVd\(~J 1 . . (second stage)
But from the above, and since P c = P<i,
which is the general expression for work of a twostage compressor without
clearance, perfect intercooling, and may be restated with the usual symbols as 
follows :
W = 144~ T (sup.pr.)(L. P. Cap.) (R P i)+(R p2 y2 , . (79)
L J
n which (R P i) and R P z) are the ratios of delivery to supply pressures for the j
irst stage and for the second stage respectively. From Eq. (79), work per i
jubic foot of gas supplied is,
W s f s ~ 1 s ~ 1 1
"~~ 2 ' ' (80)
Work per cubic foot of gas discharge and cooled to its original temperature is
w _ s r s i tz. 1 "I
(H. P. Cap. cold) 81
(81)
The lowpressure capacity stated in terms of highpressure capacity hot,
is actually discharged is
i_
(L. P. Cap.) = (H. P. Cap. hot)# p2 s R P i, ..... (82)
whence

per cubic foot hot gas discharged
(H. P. (Tap.
2 . (83)
WORK OF COMPRESSORS
105
Examination of Fig. 28 will show without analysis that there must be some
bitreceiver pressure at which least work will be required. For if the receiver
p ; ssure approached PD then the compression would approach single stage and
II
looj saunbs aod spunod uj saansseaa:
je compression line approach BCG. The same would be true as the receiver
jessure approached P g = P e , whereas at any intermediate point C, intercooling
tuses the process 'to follow BCDE with a saving of work over singlestage
oeration represented by the area DCGE. This area being zero when C is at
106 ENGINEERING THERMODYNAMICS
either B or G, it must have a maximum value somewhere between, and
pressure at which this leastcompressor work will be attained is the bcstrecei
pressure.
By definition the bestreceiver pressure is that for which W is a minimu
or that corresponding to
dp,"
Performing this differentiation upon Eq. (78), equating the result to ze
and solving for P c ,
(Best rec.pr.) = (P b P e )*~ [(sup. pr.) (del. pr)]* (*;
Substituting this value in the general expression for work Eq. (78), noting tl
Eq. (85) is the general expression for twostage work with perfect int
cpoling at bestreceiver pressure in terms of pressures and volumes. Sn
stituting the symbols for the pressures and volumes and noting that aso
Cycle 1,
F & =(L. P. Cap.) and F e =(H. P. Cap. hot) and using; (#) for (~\
= 288(sup.pr.)(L. P. Cap.)
This equation gives the same value as Eq. (85), but in terms of different u
It should be noted here that the substitution of bestreceiver pressure in
expressions for the two stages preceding Eq. (78), will show that the work d
in the two cylinders is equal.
The work per cubic foot of lowpressure gas, from Eq. (86) is,
(L.pcap.)
To transform Eq. (85) into a form involving delivery volumes, use the r
tion from the diagram,
i
pj\p..
WORK OF COMPRESSORS 107
Whence
e \p b j\p d ,
which for the bestreceiver pressure becomes
8+1
Substituting in Eq. (85),
Introducing the symbols,
. P. Cap. hot)E p ~2s~ R p ~lte~ 1 , . . (89)
and
s1
 ' ' (90)
The volume of gas discharged at the higher pressure when reduced to its
original temperature will become such that
(L.P.Cap.) = P e
(H. P. Cap. cold) P b py
or
(sup.pr.) (L. P. Cap.) = (del.pr.) (H. P. Cap. cold), . . . (91)
which may be substituted in Eq. (86),
.pr.)(H.P. Cap. cold)/2rl . . . (92)
from wrrich the work per cubic foot of gas delivered and cooled is,
w
(H.P. Cap.
108 ENGINEERING THEEMODYNAMICS
Example 1. Method of calculating diagram, Fig. 28.
Assumed data:
Fa =0 cu.ft. P a =2116 Ibs. per square foot.
F/=0 cu.ft. P c =P d = VP a P e =6172 Ibs. sq.ft.
F 6 = 5 cu.ft. P/ = P.=P,= 18,000 Ibs. sq.ft.
5 = 1.4.
To obtain point C:
or
/. F c = 2.36 cu.ft. P c = 6172 Ibs. sq.ft
To obtain point D:
.*. . F d = 1.71 cu.ft. Po = 6172 sq.ft.
To obtain point E:
but by definition
hence,
Fe = 1.71 2.14 = .8 cu.ft. P e = 18000 Ibs. sq.ft.
Example 2. To compress 5 cu.ft. of air from one atmosphere (2116 Ibs. per sq\
foot) to 8.5 atmospheres (18,050 Ibs. per square foot) in two stages with bestreceive
pressure and perfect intercooling requires how much work?
W =288(sup.pr.)(L. P. Cap.)(fl p 1),
s 1
(sup.pr.) =14.7. (L. P. Cap.) =5. fl^8.5.
/ irJL \
.'. TF=288X3.463X14.7X5X (8.5 2a ij =26,800 ft.lbs.
WORK OF COMPRESSORS 109
Prob. 1. Air at 14 Ibs. per square inch absolute is compressed to 150 Ibs. per square
nch absolute by a twostage compressor. What will be the work per cubic foot of air
elivered? What will be the work per cubic foot if the air be allowed to cool to the
riginal temperature, and how will this compare with the work per cubic foot of sup
ilied air? Best receiverpressure and perfect intercooling are assumed for the above
ompressor, s = 1.4.
Prob. 2. A compressor receives air at atmosphere and compresses it to half its vol
.me, whereupon the air is discharged to the cooler and its temperature reduced to the
riginal point. It then enters a second cylinder and is compressed to 80 Ibs. absolute.
Vliat will be the work per cubic foot of supplied air in each cylinder and how will
he work of compressing a cubic foot to the delivery pressure compare with the work
.one if compression were single stage, compression being adiabatic.
Prob. 3. Air is to be compressed from 15 Ibs. per square inch absolute to 10 times
jhis pressure. What would be the bestreceiver pressure for a twostage compressor?
v flow many more cubic feet may be compressed per minute in two stage than one stage
iy the same horsepower?
Prob. 4. A manufacturer sells a compressor to run at bestreceiver pressure
tfhen (sup.pr.) is 14 Ibs. per square inch absolute and (del.pr.) 100 Ibs. per square inch
1 .bsolute. What will be the work per cubic foot of supplypressure air done in each cylin
j,ier? Another compressor is so designed that the receiver pressure for same supply
pressure and delivery pressure is 30 Ibs. per square inch absolute, while a third is so
lesigned that receiver pressure is 50 Ibs. per square inch absolute. How will the
rork done in each cylinder of these machines compare with that of first machine?
Prob. 5. For an ice machine a compressor works between 50 and 150 Ibs. per square
ach absolute. It is single stage. Would the saving by making compression two stage
t bestreceiver pressure amount to a small or large per cent of the work in case of single
tage, how much?
Prob. 6. A compressor has been designed to compress 1000 cu.ft. of carbon dioxide
ter minute from 15 to 150 Ibs. per square inch absolute. What horsepower will be
;equired at bestreceiver pressure? Should delivery pressure change to 200 Ibs., what
ower would be required? To 100 Ibs. what power?
Prob. 7. A gascompressing company operates a compressor which has to draw
00 2 gas from a spring and compress it to 150 Ibs. per square inch gage. In the morning
ressure on the spring is 10 Ibs. gage, while by evening it has dropped to 5 Ibs. absolute.
pf the compressor was designed for the first condition, how will the highpressure
Capacity cold and horsepower per cubic foot of highpressure gas at night compare
Sirith corresponding values in morning? Assume a barometric reading.
Prob. 8. On a mining operation a compressor is supplying a number of drills and
moists with air at 150 Ibs. per square inch absolute, the supply pressure being 14 Ibs.
ijjVhat will be the difference in horsepower per cubic foot of delivered air at compressor
jud per cubic foot received at drills if air is a long time in reaching drills?
Prob. 9. With a bestreceiver pressure of 40 Ibs. per square inch absolute and a
upply pressure of 14 Ibs. per square inch absolute, what horsepower will be required
1 o compress and deliver 1000 cu.ft. of highpressure air per minute at the delivery
>ressure for which compressor is designed and what is that delivery pressure?
10. Twostage Compressor, with Clearance, Perfect Intercooling Expo
icntial Compression, Bestreceiver Pressure, Equality of Stages, (Cycle 6).
Work and Capacity in Terms of Pressures and Volumes. The twostage expo
110
ENGINEERING THERMODYNAMICS
nential compressor with clearance and perfect intercooling is represented by
the PV diagrams Figs. 29, 30, 31, which are clearly made up of two singlestage
compression processes, each with clearance.
^ooj 94Bnbg aad sqi ui seanssaaj; w
I
a
Applying Eq. (57) to the two stages and supplying proper subscripts;
referring to Fig. 29,
81
F a )[^ ' l]. .
(first stage)
WORK OF COMPRESSORS 111
*z_ J
1 . . (second stage)
[f the condition of" perfect intercooling be imposed, it is plain that since
n weight of gas entering the second stage must equal that entering the first
jj;e, and the temperature in each case is the same,
jj noting also that
(94)
B (94) is the general expression for the work of twostage exponential
:cipressor with perfect intercooling, P c being the receiver pressure.
p
As in Section 9, let (R P i) be the pressure ratio  for the first stage and(72?2)
"b
Pe
ih pressure ratio for the second stage and using instead of P& its equivalent
PC
A (sup.pr.) Ibs. per square inch.
1 1 . . (95)
fjch is identical with (79), showing that for twostage compressors with perfect
nrcooling (as for single stage, Section 7), the work for a given lowpressure
aicity is independent of clearance.
The work per cubic foot of gas supplied is given by Eq. (80); per cubic
0) of cold gas delivered by Eq. (81) and per cubic foot hot gas delivered by
I (83).
The reasoning regarding bestreceiver pressure followed out in Section 9, will
dW
1 again in this case, and by putting = in Eq. (94), and solving for P c
dP c
all again be found that bestreceiver pressure will be
(bestrec. pr.) = (P 6 P e )* ......... (96)
Substitution of this value for P c in Eq. (94), gives the following expression
Dvork of the twostage exponential compressor with bestreceiver pressure,
(97)
112
ENGINEERING THERMODYNAMICS
which may be expressed in terms of supply pressure, pounds per square in
lowpressure capacity, cubic feet, and ratio of compression,
^ (sup.pr.) (L. P. Cap.) \R P '2. *  1 J ,
which is the same as Eq. (86).
Substitution of the value of best receiver pressure in the expression for
WORK OF COMPRESSORS 11^
Orkof the two stages separately will show the equality of work done in the
ispective stages for this case with clearance.
r ork per cubic foot gas supplied to compressor is
w
(gg)
ork per cubic foot of highpressure gas hot is
W s lir irJ 1
288 ^!^^*' 2s I* "
The work per cubic foot of air delivered and cooled to its original tem
iture is,
w 9 r ~ i i
(H. P. Cap. cold)  ^^(deLpr.) ^~ 1 J ,  . (101)
Due to the fact that clearance has no effect upon the work per cubic foot
substance, as previously noted, Eqs. (99), (100) and (101) are identical with
*7), (90) and (93).
11. TwoStage Compressor, any Receiver Pressure, Exponential Compres
lion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and
Eorsepower, in Terms of Dimensions of Cylinders and Clearances. Referring to
frig. 29, let DI be the displacement of the first stage cylinder in cubic feet =
Vb Vt), D2 the displacement of the cylinder of the second stage in cubic
lpet = (Fd Vf)j ci the clearance of the first stage, stated as a fraction of the
Displacement of that cylinder, so that the clearance of the first stage cubic
9et = ciDi, and that of the second stage = c^Di.
The lowpressure capacity of the first stage (L. P. Cap.) in cubic feet is
VbVa), and, as for the singlestage compressor, is expressed in terms of dis
placement, clearance and ratio of compression of the first stage as follows, see
"Sq. (64);
(L.P. Cap.i)=Dil+cici# P i =DiE vl ...... (102)
For the second stage, the lowpressure capacity (L. P. Cap. 2) is
jjnd is equal to
I
(L.P. Czp.2)=D 2 l + c 2 C2Rp2 s =D 2 E C2 .... (103)
Volumetric efficiency of the first stage is given by
i
* ....... (104)
t I
I
metric efficiency for second stage
ENGINEERING THERMODYNAMICS
(105
I
6
I
02
6
2
It may be required to find the receiver pressure (incidental to the finding c ;
work or horsepower) for a compressor with given cylinder sizes and deliver!
pressure. The condition assumed of perfect intercooling stipulates that
(L. P. Cap.i)(sup.pr.) = (L. P. Cap. 2 ) (rec.pr.),
WORK OF COMPRESSORS
'hence
.,(L.P.Cap.i)
(106)
If the volumetric efficiencies are known or can be sufficiently well approx
.iiated this can be solved directly. If, however, E v \ and E& are not known,
ut the clearances are known, since these are both dependent upon the receiver
ressure sought, the substitution of the values of these two quantities will give
(rec.pr.) = (sup.pr.)  ^, . . . (107)
D2\ l~f"C2 C2\ '
\ rec.pr.
,n expression which contains the receiver pressure on both sides of the equation.
: f his can be rearranged with respect to (rec. pr.), but results in a very complex
impression which is difficult to solve and not of sufficient value ordinarily to
varrant the expenditure of much labor in the solution. Therefore, the relations
ire left in the form (107). It may be solved by a series of approximations,
,he first of which is
(rec.pr.) = (sup.pr.)^ approx. ...... (108)
L/2
Vith this value for the receiver pressure, substitution may be made in the sec
ond member of the Eq. (107), giving a result which will be very nearly correct,
if desirable, a third approximation could be made.
To find the work of a twostage exponential compressor in terms of displace
ment of cylinders, supply pressure, receiver pressure and delivery pressure,
>ounds per square inch, and volumetric efficiency of the first stage, E v i, from (79)
r (94) ,
. . . ,(109)
a J. L J
in which
(rec.pr.) (del.pr.)
v f ~A p \.
(sup.pr.) (rec.pr.)
,Fo solve this the receiver pressure must be found as previously explained and
;he volumetric efficiency must be computed by Eq. (104) or otherwise be known.
It is impracticable to state work for this general case in terms of displace
ment and clearances directly, due to the difficulty of solving for (rec. pr.) in Eq.
116 ENGINEERING THERMODYNAMICS
(107). It may, however, be stated purely in terms of supply and delivery pres
ejures, in pounds per square inch, displacement, in cubic feet, and volumetri
efficiencies, as follows:
From Eq. (106),
and
_del.pr. ..D 2 E v2 _ D 2 E v2
Kp2 ~ sup.pr. X Z>i#,i ~ ^DiEn'
Hence
The mean effective pressure of the twostage compressor referred to the lowl
pressure cylinder is found by dividing the work of the entire cycle Eq. (110), b
the displacement of the firststage cylinder, and by 144, to give pounds pe
square inch.
m.e.p. referred to firststage cylinder, pounds per square inch is,
It is well to note that this may also be found by multiplying (work don:
per cubic foot of gas supplied) by (volumetric efficiency of the first stage, E v \]
and dividing the product by 144.
In terms of the same quantities, an expression for indicated hbrsepowe
may be given as follows:
! a1
s (sup.pr.)  p v .
f f, ~ 2 ,
1
~ 2J ,
where n is the number of cycles completed per minute by the compressor
For n may be substituted the number of revolutions per minute, divided by th
revolutions per cycle,
N
The horsepower per cubic foot of gas supplied per minute is
'I.H.P s (sup.pr.) \(DiE vl \s / D2E v2 \* J
n(L.P.Cap.) "7^1 ^29T" [\D 2 E v2 ) f ("'D^J
WORK OF COMPRESSORS 117
; orsepower per cubic foot of gas delivered and cooled per minute.
s i .i
n(H. P. Cap. cold) sl 229.2 \D 2 E
orsepower per cubic foot of hot gas delivered per minute
 LH.P. s sup.pT.(D l E 9 i\ L ir L
i n(H.P.Cap. hot) sl 229.2 \D 2 E, 2 )
R _
"
7^_
2 
lor the case where clearance is zero or negligible, these expressions may be
mplified by putting E,? and E i equal to unity.
_
s (sup.pr.) [/DA . ,( R D 2 \ _J
~i=i"M95~ ;I LW f v w
,H.P. per cubic foot, gas supplied per minute
D\ ]
J'
n(L. P. Cap.) ~
H.P. per cubic foot gas delivered and cooled per minute
I.H.P. (del.prOr/gA?' /Da\'
n(H. P. Cap. cold)" sl 229.2
.P. per cubic foot hot gas delivered per minute
Example 1. Method of calculating diagram, Figs. 29, 30, 31.
\(.Sf>umed data.
p a =p b =2116 Ibs. per square foot;
p c = p d =p h =p k =:Ql72 Ibs. per square foot.
P g =P e =P f = 18,000 Ibs. per square foot.
7(H. P.) = 7.5 per cent; C7(L. P.) 7.5 per cent; s = 1.4; L. P. Capacity =5 cu.ft,
118 ENGINEERING THERMODYNAMICS
To obtain point A".
From formula Eq. (64),
i
5 = Di(l + .075.075X2.14), hence Di =5.45 cu.ft.
Cli =V k = 5.45 X .075 = .41 cu.ft.
/. Ft = .4 cu.ft.; P t =6172 Ibs. sq.ft.
To obtain point A :
i
F a = F*(^Y' 4 = .4X2.14 = .856 cu.ft.
\* </
.'. Fa =.85 cu.ft.; P a =2116 Ibs. sq.ft.
To obtain point B:
F 6 = F +5 = .85 +5 =5.85 cu.ft.; P 6 =2116. Ibs. sq.ft.
To obtain point C:
' =5.85 s2.14 =2.73,
4
F c =2.73 cu.ft.; P c = 6172 Ibs. sq.ft.
To obtain point D:
Volume at D is the displacement plus clearance of H. P. cylinder. This cannot I
found until the capacity is known. The capacity is the amount gas which must b
taken in each stroke and which is also the amount actually delivered by L. P. cylindc
cooled to original temperature. The amount of cool gas taken in by the second cylind
is
L V
(L. P. Cap. 2 ) = ~(L P. Cap.O = X5 = 1.7 cu.ft.
But
cu.ft.
. 14=2.02 cu.ft.
P d = 6172 Ibs. sq.ft.
Other points are easily determined by relations too obvious to warrant setting dow:
Example 2. What will be the capacity, volumetric efficiency and horsepower p
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow j
ing compressor: Twostage, doubleacting cylinders, 22 and 34jX24in., runnirf
at 100 R.P.M. Highpressure clearance 6 per cent, lowpressure 4 per cent. Supp i
WORK OF COMPRESSORS 119
essure 14 Ibs. per square inch absolute. Delivery pressure 115 Ibs. per square inch
>solute.
The capacity will be the cylinder displacement times the volumetric efficiency.
Pi = displacement of a 34ix24 in. cylinder, or 12.8 cu.ft. and Z>2 = displacement of
22+24 /r cylinder or 5.4 cu.ft. To obtain the volumetric efficiency, make use of
;>proximation of formula Eq. (108),
D 12 8
(rec.pr.) = (sup.pr.) =14 X~r =33.2 Ibs. sq.in,
L/z o.4
id then by Eq. (107) checking,
(rec.pr.) =14X i^ ^ ' =35.1 Ibs. sq.in.
__
E n = 1 +ci Ci(R pl ) s from Eq. (104),
i
= l+.04.04x(2.5)~ s =96.8 per cent
herefore the capacity will be,
200 X 12.8 X. 968 =2480 cu.ft. per minute;
E C2 = 1 +c 2 C 2 (# P2 ) from Eq. (105),
= 1+.06.06X(3.28)' 714 =92 percent.
from Eq. (113), I H.P. per cu.ft. (sup.pr.) air per minute is,
sl sl
s sup.pr. [ (DJti\ ( K D 2 E V2 \ . ]
= rl "229T L \D 2 E~J f V*KR/,
1.4 14 r/12.8x.968\286 / ^4x^\.286 n
T X 229^2 L \^4XlJ2"/ 22 12.8X.968/
/"hence horsepower per 1000 cu.ft. of free air per minute is, =150.
From Eq. (115) horsepower per cubic foot (del.pr.) air, hot = that of (sup.pr.)
( L  j or 5.85 times that of (sup.pr.) air.
,'. Horsepower per 1000 cu.ft. of hot (del.pr. air) = 150 X5.85 =877.
i Prob. 1. A twostage doubleacting compressor has volumetric efficiencies as shown
y cards of 98 per cent and 90 per cent for the high and lowpressure cylinders respect
/ely. It is running at 80 R.P.M. and compressing from atmosphere to 80 Ibs. per square
ich gage. If the cylinders are 15jx25i Xl8 ins., and speed is 120 R.P.M., what
120 ENGINEERING THERMODYNAMICS
horsepower is being used and how many cubic feet of free and compressed air (hoi
and cold) are being delivered per minute, when s equals 1.41?
Prob. 2. What horsepower will be needed to drive a twostage compressor 10 \ ins
and 16i Xl2 ins., double acting, with 5 per cent clearance in each cylinder at 160 R.P.M
when the supply pressure is atmosphere, delivery pressure 100 Ibs. per square incl
gage, when s 'equals 1.35?
Prob. 3. A thousand cubic feet of free air per minute must be compressed by a two
stage compressor to 80 Ibs. per square inch gage from a supply pressure of 10 Ibs. pei
square inch absolute. The volumetric efficiencies for the high and lowpressure cylin
ders are 85 per cent and 95 per cent respectively, and the receiver pressure is 25 Ibs
per square inch absolute. What will be the displacement of each cylinder and th<
horsepower per cubic foot of (sup.pr.) air?
Prob. 4. How many cubic feet of free air can be compressed in twostage compres
sor 18ix30i X24 ins. with 5 per cent clearance in highpressure cylinder and 3 per cem
in low if (sup.pr.) is atmosphere and (del.pr.) 80 Ibs. per square inch gage? How woulc
the answer be affected if clearance were taken as zero? Take s = 1.41.
Prob. 5. The volumetric efficiency of the lowpressure cylinder is known to be 95 pe,
cent, and of the highpressure cylinder 85 per cent. The cylinder sizes are 15 X25^ xli
ins. and speed is 120 R.P.M. What horsepower must be supplied to the machine ij
the mechanical efficiency is 80 per cent and the pressure ratio 10 with a (sup.pr.) of on
atmosphere?
Prob. 6. A compressor runs at 120 R.P.M. and is double acting. It is compressing
air from 14 Ibs. per square inch absolute to pressures ranging from 70 Ibs. per squar<
inch gage to 100 Ibs. per square inch gage. The cylinders are 20}x32ix24 ins., am
clearances 8 per cent and 4 per cent. Find the approximate receiver pressure, capacity
and horsepower for the range of discharge pressure, for s = 1.3.
Prob. 7. The volumetric efficiency of the lowpressure cylinder of a twostage com
pressor is known to be 95 per cent, the receiver pressure as shown by gage is 40 Ibs.
delivery pressure 100 Ibs., and supply pressure one atmosphere. What will be th>
horsepower if the machine runs at 120 R.P.M. and the lowpressure cylinder is 18 X 12 in.
s = 1.4.
Prob. 8. An air compressor appears to require more power to run it than shouL
be necessary. It is a doubleacting 18x30x24 in. machine running at 100 R.P.M
The volumetric efficiencies are 85 per cent and 90 per cent respectively and supply an<
delivery pressures 14 Ibs. and 110 Ibs. per square inch, both absolute. What would b f
the minimum work per cubic foot of (sup.pr.) air, per cubic foot of (del.pr.) air, hot an</.
cold, for adiabatic compression?
Prob. 9. The efficiency of the driving gear on an electricdriven compressor i^
75 per cent. Power is being supplied at the rate of 150 H.P. How much air shoul*;
be compressed per minute from 4 Ibs. per square inch absolute to 100 Ibs. per squarl
inch gage, if the receiver pressure is 35 Ibs. per square inch absolute and the lowpressur i
volumetric efficiency is 90 per cent, s being 1.4?
12. Twostage Compressor with Best Receiver Pressure Exponents
Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pres
sures and Horsepower in Terms of Dimensions of Cylinders and Clearances
For the twostage exponential compressor with or without clearance, and pei
feet intercooling, the bestreceiver pressure was found to be (Eq. 84),
(bestrec.pr.) = [(sup.pr.) (del.pr.)]* (12C
WORK OF COMPRESSORS 121
This expression Eq. (120) for bestreceiver pressure makes it possible to
aluate R P i and R p2 as follows:
sup.pr. sup.pr. L su PP r
(del.pr.) (del.pr.)
for (bestrec. pr.) = 77 >. = r f w , , . u
(bestrec.pr.) [(sup.pr.) (del.pr.)]'
r/deLprAl t =
L \sup.pr.y J
The use of these values for R p \ and R P 2 in the expressions previously given
}>r volumetric efficiency for the general case, Eqs. (104) and (105) results in
iolumetric efficiency, first stage
_i
ad volumetric efficiency, second stage
i_
i
The work was found to be represented by Eq. (98), which may be stated
terms of displacement and volumetric efficiency of the first stage, as follows:
.... (125)
here R p = ( G ' pr \ and where (sup.pr.) is in pounds per square inch,
(sup.pr.;
If the clearance is known for the first stage this becomes by the use of
q. (104),
2rl, .. . (126)
s
Kich is a direct statement of the work of a twostage adiabatic compressor
ith perfect intercooling in terms of supply pressure and delivery pressure,
ounds per square inch, displacement, cubic feet and clearance as a fraction
122 ENGINEERING THERMODYNAMICS
of displacement, provided the cylinder sizes and clearances are known to be sud
as to give bestreceiver pressures.
The mean effective pressure reduced to firststage displacement, in pounds
per square inch, may be derived from either Eq. (125) or (126) by dividing tl
work by the displacement of the firststage cylinder, and again dividing by
2s / _L\ r L~ 1
= ^ (sup.pr.)l l+cici/^2* }\R P 2 S 1
(127.
Since the work done is equally divided between the two cylinders when best
receiver pressure is maintained, the mean effective pressure, in pounds pei
square foot, for each cylinder will be, onehalf the total work divided by th(
displacement of the cylinder in question,
w s r s ~ i i
m.e.p., first stage = 2gg^= j(sup.pr.)^i /^ 2. l. ..... (128
Note that this is onehalf as great as the m.e.p. of the compressor reducec
to first stage, (127),
w s , N Di r ^
m.e.p., second stage = 2ggyT = ^2Y(sup.pr.)^ 7 P iLR 1? 2s 1 1 . . . (129
But
D\E c i [ 1
(sup.pr.)^ ^ = (rec.pr.) = (sup.pr.)(del.pr.) p,
Lf2^v2
whence,
s If 1 f 1
.e.p., second stage= (sup.pr.)(del.pr.) U^LRp 2 1. . . . (II
m
It is next necessary to investigate what conditions must be fulfilled to obtai
the bestreceiver pressure, the value of which is stated, Eq. (120). The conditiol
of perfect intercooling provides that the temperature of the gas entering tr
second stage is the same as that entering the first stage, and hence that tl !
product (volume entering second stage) X (pressure when entering second stage '
must be equal to the product (volume entering first stage) X (pressure of suppl
to first stage), or
(L. P. Cap. 2 )(rea pr.) = (L. P. Cap. i) (sup. pr.), . . . (12
WOEK OF COMPEESSOES 123
Ombining with Eq. (120)
(L. P. Cap.Q [(sup.pr.) (del.pr.)]* [ (del.pr.) ]*_ .
(L. P. Cap. 2 ) ~ (sup.pr.) [ (sup.pr.) J "
(1) (2) r (3) J,
^_(L.P.C*p. 1 )_D l E vl _ Dl [ 1 + Cl  ClR ^\
r i * * * *
: 1 4~ C2 C2^p2s I
(L. P. Cap. 2 ) *,* D
un this threepart equation proper values may be found to fulfill require
.ents of bestreceiver pressure for :
1. The ratio of capacities for a given ratio of pressures, or conversely, the
ratio of pressures when capacities are known;
! 2. The ratio of cylinder displacements for known volumetric efficiencies;
3. The ratio of cylinder displacements when the clearances and ratio of com
ression are known, or conversely, with known displacements and clearances
jiie ratio of pressures which will cause bestreceiver pressure to exist. This
ist case in general is subject to solution most easily by a series of approxi
lations.
There is, however, a special case which is more or less likely to occur in prac
ice, and which lends itself to solution, that of equal clearance percentages. If
i = C2 the parenthesis in the numerator of part (3) of Eq. (132) becomes equal
D the parenthesis in the denominator, and evidently the volumetric efficiency
if the two cylinders are equal, hence for equal clearance percentages in the two
ages,
 ......
f case which leads to the same expression, Eq. (133), is that of zero clearance,
I condition that is often assumed in machines where the clearance is quite
LThe work per cycle, Eq. (126), when multiplied by the number of cycles
rformed per minute, n, and divided by 33,000, gives
l), (134)
from which are obtained the following:
[.H.P. per cubic foot supplied per minute
n(L.P. Cap.) 81 114.6
f .H.P. per cubic foot delivered and cooled per minute
I.H.P. s (del.pr.)
n(H. P. Cap. cold) s1 114.6 (
124 [ENGINEERING THERMODYNAMICS
and I.H.P. per cubic foot delivered hot per minute
I.H.P. . (sup.pr.) +! =!
2s (K p 2s 1). . (137.
n(H. P. Cap. hot) s 1 114.6
These expressions, Eqs. (165), (166) and (167) are all independent of clear
ance.
Example. What will be the capacity, volumetric efficiency and horsepower pe
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for th
following compressor for s = 1.4? Twostage, doubleacting, cylinders 22 X34^ X24 ins.
running at 100 R.P.M. Lowpressure clearance 5 per cent, highpressure clearanc
such as to give bestreceiver pressure. Supply pressure 15 Ibs. per square inch abso
lute, delivery pressure 105 Ibs. per square inch absolute.
Capacity will be cylinder displacement times low pressure volumetric efficiency, ori
200Z>iX#,i.
Di = 17.5 cu.ft.
En from Eq. (123)  (1 +Ci dR P 2~s)
= 1+.05.05X7' 357 =95 per cent.
Therefore low pressure capacity = 200 X 12.8 X. 95 =2430 cu.ft. per minute.
Horsepower per cubic foot of (sup.pr.) air per minute is from Eq. (135)
.<? siirt rr * 1
1
Therefore, horsepower per 1000 cu.ft. of sup.pr. air = 160.
Horsepower per cubic foot of (del.pr.) air, hot, is from Eq. (137)
1+8
R p 2s times power per cu.ft. of (sup.pr.) air,
hence,
160 X5.3 =850 =horsepower per 1000 cu.ft. of (del.pr.) air, hot, per minute.
Problem Note. In the following problems, cylinders are assumed to be proportione
with reference to pressures so as to give bestreceiver pressure. Where data conflic
the conflict must be found and eliminated.
Prob. 1. Air is compressed adiabatically from 14 Ibs. per square inch absolute t
80 Ibs. per square inch gage, in a 20j X32 X24 in. compressor, running at 100 R.P.M
the lowpressure cylinder has 3 per cent clearance. What will be horsepower J:<
quired, to run compres or and what will be the capacity in cubic feet of low prcssu:
and in cubic feet of (del.pr.) air?
Prob. 2. What must be the cjdinder displacement of a twostage compressor with
per cent clearance in each cylinder to compress 500 cu.ft. of free air per minute fro)
14 Ibs. per square inch absolute to 85 Ibs. per square inch gage, so that s equals I/,
What will be the horsepower per cubic foot of (del.pr.) air hot and cold?
Prob. 3. A twostage compressor is compressing gas with a value of s = l.
from 10 Ibs. per square inch gage to 100 Ibs. per square inch gage. The cylinders ai
18^X301x24 ins., and speed is 100 R.P.M. If the clearance is 5 per cent in the lov
WOEK OF COMPRESSORS 125
I assure cylinder and 10 per cent in the high, what will be the cubic foot of (sup.pr.) gas
i ndled per minute and what will be the horsepower at best receiver pressure?
Prob. 4. A manufacturer states that his 20 jX 32^x24 in. doubleacting compres
gr when running at 100 R.P.M. at sea level will have a capacity of 2390 cu.ft. of free air
jr minute, pressure range being from atmosphere to 80 Ibs. per square inch gage. At
tstreceiver pressure what clearance must the compressor have, compression being
i iabatic?
Prob. 5. The cylinder sizes of a twostage compressor are given as 10ixl6xl2
13., and clearance in each is 5 per cent. What will be the bestreceiver pressures when
(i crating between atmosphere and following discharge pressures, 60, 70, 80, 90, 100
td 110 Ibs. per square inch gage, for s equal 1.4?
Prob. 6. 1500 cu.ft. of air at 150 Ibs. per square inch gage pressure are needed per
inute for drills, hoists, etc. The air is supplied from 3 compressors of the same size
(d speed, 120 R.P.M. Each has 4 per cent clearance in each cylinder. What will
! sizes of cylinders and the horsepower of the plant for bestreceiver pressure, when
JM.41?
Prob. 7. The cards taken from a compressor show volumetric efficiencies of 95 per
mt and 80 per cent .in low and highpressure cylinder respectively. What will be
iel.pr.) for bestreceiver pressure if compressor is 15lx25jXl8 ins., and (sup.pr.) 15
jk per square inch absolute to 10 Ibs. absolute, and what will be the work in each case,
being 1.35?
Prob. 8. A manufacturer gives a range of working pressure of his lOjx 16^X12 in.
<mpressor from 80100 Ibs. per square inch page. If clearances are, low 4 per cent,
Ugh 8 per cent, and (sup.pr.) is atmosphere, find by trial which end of the range comes
Barest to giving bestreceiver pressure? If clearances were equal which would give
listreceiver pressure?
Prob. 9 A 16 J X25j Xl6 in. compressor is rated at 1205 cu.ft. free air per minute at
;i5 R.P.M. at sea level. What would be the clearance if compressor were compressing
a* from atmosphere to 100 Ibs. gage at sea level? With same clearance what would be
ie size of a lowpressure cylinder to give the same capacity at altitude of 10,000 ft. with
e same clearance and the same (del.pr.) , bestreceiver pressure always being maintained?
13. ThreeStage Compressor, no Clearance, Perfect Interceding Expo
sntial Compression (Cycle 7), Best Two Receiver Pressures, Equality of
;tages. Work and Capacity, in Terms of Pressures and Volumes. The three
age exponential compressor cycle with no clearance, perfect intercooling Cycle
il is shown in Fig. 32. The net work area, ABCDEFGHJKA , is made up of
Iliree areas which may be computed individually by the formulae for single stage
i?q. (48), provided the requisite pressures and volumes are known, as follows:
1
W = ~ri P V [(j?) *  1 ] (first stage)
sl
~Tl P V [(j?) S ~ l ] (second stage)
s l
(third stage)
. . (138)
126 ENGINEERING THERMODYNAMICS
But the condition of perfect intercooling provides that for no clearance,
Q
&&
(JdJoaa)
m
\q; aa^nbg jsd 'sc
and it may be noted that Pd = P C) and P f =P e . Accordingly,
sl s1 f s1
Pressures in this expression are in pounds per square foot.
. . (14
WORK OF COMPRESSORS 127
Changing the equation to read in terms of supply pressure pounds per square
ich, lowpressure capacity cubic feet, and ratios of pressures, first stage
, second stage (R P 2) and third stage (R P z), it becomes
/ork done by threestage compressor, perfect intercooling
r si si 91
.P. Cap.) \(R p i) ^~ + (R v 2)~* +(Rpz) 31, (141
this the following expressions are derived:
rk per cubic foot supplied
si sl * i
3]. . . (142)
(143)
rk per cubic foot gas delivered and cooled
1 3] .
ork per cubic foot gas, as delivered hot
sl
(144)
Best Two Receiver Pressures. Referring to Fig. 32, P c is the pressure in the
fjst receiver (1 rec.pr.) and P e is the pressure in the second receiver, (2 rec.pr.).
3 is evident that if either receiver pressure be fixed and the other is varied,
ie work necessary to compress a given initial volume of gas will be varied,
j&d will have a minimum value for some particular value of the varying receiver
f {essure. By a variation of both receiver pressures a minumim may be found
jlr the work when both receiver pressures have some specific relation to supply
:b.d delivery pressures. For instance, assume that P c is fixed. Then a change
i P e can change only the work of the second and third stages, and the three
eige compressor may be regarded as consisting of
One singlestage compressor, compressing form P& to P c .
One twostage compressor, compressing from P c to P .
. In this twostage compressor, bestreceiver pressure is to exist, accord
i. to Eq. (84),
P e = (best 2 rec .pr.) = (P C P,)*. . . . (145)
128 ENGINEERING THERMODYNAMICS
Similar reasoning, assuming P e fixed and making P c variable, would shoi
that
P c =(best 1 rec.pr.) = (P e P 6 )*. ...... (146
Eliminate P c from Eq. (145) and the expression becomes,
P e = (best 2 rec.pr.) = (P,P 2 ) = [(sup.pr.Xdel.pr.) 2 ] * (1
Similarly, from Eq. (146)
P c = (best 1 rec.pr.) = (P, 2 P g ) = [(sup.pr.) 2 (del.pr.)] * (14*
From these expressions may be obtained,
p. p p
z& r c r e
or
Rpi =Rp2 R
Substitution in Eq. (140) gives,
Work, threestage, bestreceiver pressure no clearance
Arranging this equation to read in terms of supply pressure, pounds per squa .
inch, lowpressure capacity, cubic feet, and ratio of pressures
Work, threestage bestreceiver pressure
sl
= 432^(sup.pr.)(L.P. Cap.)(#/37 1), . . . (u
s i
The work of the compressor is equally divided between the three staty
when bestreceiver pressures are maintained, which may be proven by substi
tion of Eq. (149) in the three parts of Eq. (138), and
Work of any one stage of threestage compressor with bestreceiver pressure
sl
WORK OF COMPRESSORS 129
From Eq.(151), may be derived the expressions for work per unit of capacity.
\Drk per cubic foot lowpressure gas is,
(L. P. Cap.) = (H. P. Cap. co\d)R p .
lork per cubic foot cooled gas delivered is,
(H. P. Cap. cold) s1
/rain, from Fig. 32,
'hich is to say that, when bestreceiver pressures are maintained,
(L. P. Cap.) = (H. P. Cap.
Example 1. Method of calculating Diagram, Fig. 32.
ssumed data.
P a =P b = 2llQ Ibs. per sq.ft.
PC =Pa =best firstroceiver pressure =P a P g * =4330.
P e =P/=best secondreceiver pressure P^P g ^ =8830.
P a =P h = 18,000 Ibs. per sq.ft.
7&=5cu.ft. s = 1.4.
(153)
r.)# p (#/37 1 l). . . . (154)
=(H.P.Cap.hot)j%,~lT, .... (155)
bnce
fork per cubic foot hot gas delivered
W 9 2s +1 f1
(H.P. Cap. '
130 ENGINEERING THERMODYNAMICS
To obtain point C:
/. V c =3 cu.ft. P c =4330 Ibs. sq.ft.
Intermediate points B to C may be found by assuming various'pressures and finding
the corresponding volumes as for V c .
To obtain point D:
F d = F 6 X^ =5 X^ =2.44 cu.ft.
Pa 4330
/. F d =2.44 cu.ft., P d =4330 Ibs. sq.ft.
To obtain point E:
T7 T . (P e \iA P e *Pa
V e = V d + ( } , but = ,
by assumption of bestreceiver pressure.
Hence V e = 2.44^1.67 = 1.46 cu.ft., an P e = 8830 Ibs. sq.ft.
Intermediate points D to E may be found by assuming various pressures and findin ;
corresponding volumes as for V e , and succeeding points are found by similar method*;
to these already used.
7, = .72, P a = 18,000,
Example 2. What will be the horsepower required to compress 100 cu.ft. of fre
air per minute from 15 Ibs. per square inch absolute to 90 Ibs. per square inch gage in
noclearance, threestage compressor if compression be adiabatic? What will be th
work per cubic foot of (del.pr.) air hot or cold?
From Eq. (153) work per cubic foot of (sup.pr.) air is,
^X15X(7 0952 1) =4500 ft.lbs.,
or
4^00 vino
H.P. for 100 cu.ft. per minute =33^ = 13  6 
From Eq. (154) work per cubic foot of (del.pr.) air cold is R p times that per cul
foot of (sup.pr.) air, or in this case is 31,500 ft.lbs.
I From Eq. (156) work per cubic foot of (del.pr.) air hot is R p 3s times that per cub
foot of (sup.pr.) air, or in this case 5.8 X45,000 =46,200 ft.lbs.
WORK OF COMPRESSORS 131
Prob. 1. What work will be required to supply 2000 cu.ft. of air at 200 Ibs. per inch
j,ge pressure if compressing is done adiabatically by threestage compressors, taking
; T at atmosphere, neglecting the clearances?
Prob. 2. A motor is available for running a compressor for compressing gas, for
lich s equals 1.3. If 60 per cent of the input of the motor can be expended on the
,r, to what delivery pressure can a cubic foot of air at atmospheric pressure be com
;essed in a zero clearance threestage machine? How many cubic feet per minute
. uld be compressed to a pressure of 100 Ibs. gage per H.P. input to motor?
Prob. 3. Two compressors are of the same size and speed. One is compressing
r so that exponent is 1.4, the other a gas so that exponent is 1.1. Each is three stage,
'hich will require the greater power to drive, and the greater power per cubic foot
, (sup.pr.) gas, and per cubic foot of (del.pr.) gas, hot, and how much more, neglect
g clearance?
Prob. 4. How will the work per cubic foot of (sup.pr.) air and per cubic foot of (del.
K) air differ for a threestage compressor compressing from atmosphere to 150 Ibs. per
,uare inch gage from a single and a twostage, neglecting clearance?
Prob. 5. A table in " Power " gives the steam used per hour in compressing air to
irious pressures single stage. A value for air compressed to 100 Ibs. is 9.9 Ibs. steam
;;r hour per 100 cu.ft. of free air. Using the same ratio of work to steam, find the value
;r the steam if compression had been threestage, zero clearances to be assumed.
Prob. 6. A 5 in. drill requires 200 cu.ft. of free air per minute at 100 Ibs. per square
ich gage pressure. What work will be required to compress air for 20 such drills if
treestage compressors are used, compared to singlestage for no clearance?
Prob. 7. What would be the steam horsepower of a compressor delivering 150
i.ft. of air per minute at 500 Ibs. per square inch pressure if compression is threestage,
jliabatic, clearance zero, and mechanical efficiency of compressor 80 per cent?
14. Threestage Compressor with Clearance, Perfect Interceding Expo
untial Compression (Cycle 8), Bestreceiver Pressures, Equality of Stages.
v ork and Capacity in Terms of Pressures and Volumes. The pressure
T )lume diagrams of the threestage compression is shown in Figs. 33, 34 and
S, on which the clearance volume and displacements, lowpressure capacity
;id highpressure or delivery capacity for hot gas are indicated.
If perfect intercooling exists, as is here assumed,
id also ^ (157)
= (V d  V l )P d = (V r  Vi)P f
(L. P. Cap.)P 6 =(H.P. Cap. cold)P,.
Apply Eq. (57) to the three stages and the entire work done is,
W = ~P>(V<,V a )[(~?^l] (first stage)
s r /p e \ s ~ i ]
2jr Pd(V d Vi) ( ~ } s 1 (second stage)
LY* <*/
f,P,(V f  V,} V pT  1 (third stage)
o 1
. . . (158)
132
ENGINEERING THERMODYNAMICS
By use of the above conditions of perfect interceding Eq. (157) thi.<
expression becomes,
(159
t
. c
bC c
5
^ooji aa^nbg aad spunoj ui saanssoaa;
in which
p
c
WORK OF COMPRESSOES
133
L terms of supply pressure, pounds per square inch, lowpressure capacity,
coic foot and ratios of pressures as above, the work of a threestage corn
p ;ssor with perfect intercooling and with clearance is
= 14 (sup.pr.)(L.P. Cap.)
(160)
hich is identical with Eq. (141), showing that clearance has no effect upon the
ork for a given capacity.
134
ENGINEERING THERMODYNAMICS
It readily follows that the work per unit of gas is independent of clearai
and hence Eqs. (142), (143) and (144), will give a correct value for the w[
f 8
^ooj arctibg aad 'sqi
CO
5V
=3
&
per cubic foot of gas supplied, per cubic feet delivered and cooled, and 1
cubic foot as delivered hot, respectively.
Since in twostage compressors the reasoning leading to the determinat c
of bestreceiver pressure applies equally well with and without clearance, it
since the value of bestreceiver pressures for threestage are found by cf
WORK OF COMPRESSORS 135
slering the threestage a combination of one and two stagecompressors, the
sue expressions for bestreceiver pressures will hold with clearance as without;
S3 Eqs. (147) and (148).
P e =(best 2 rec.pr.) = [(sup.pr.)(del.pr.) 2 ]*.
P c = (best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*.
The use of these expressions for bestreceiver pressures leads to the same
isult as for no clearance Eq. (150), except for the volumes,
r ork, threestage] bestreceiver's pressure with 'clearance
81
(161)
3 l]
'hich is stated below in terms of supply pressure, pounds per square inch low
jessure capacity, cubic foot, and ratio of compression R p ,
r ork, threestage bestreceiver pressure.
sl
(162)
'hich is identical with Eq. (151).
From this may be obtained expressions for the work per cubic foot of low
ressure gas supplied to compressor per cubic foot of gas delivered and cooled,
;id per cubic foot of gas as delivered hot from the compressor, when the re
viverpressures are best, and these will be respectively identical with Eqs.
.53), (154), and (156), in the foregoing section.
i 15. Threestage Compressor, any Receiverpressure Exponential Com
ession. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure,
id Horsepower in Terms of Dimensions of Cylinders and Clearances.

DI = displacement of the firststage cylinder, r in cubic feet=(F& F OT );
D 2 = displacement of the secondstage cylinder, in cubic feet = (F d Ft);
Da = displacement of the thirdstage cylinder, in cubic feet = (F/ Fa).
ci, C2, cs are the clearances of the first, second and third stages respectively,
;ated as fractions of the displacement, so that,
Clearance volume, 1st stage, in cubic f eet = V m
Clearance volume, 2d stage, in cubic feet = Ft =
Clearance volume, 3d stage, in cubic f eet =Vn =
136 ENGINEERING THERMODYNAMICS
The lowpressure capacity of the first stage, and hence for the compress^
is (F 6 Fa), and in terms of clearance, ci, and displacement DI of the fir
stage is, according to Eq. (64),
(L.P.
For the second stage, the lowpressure capacity is (Va Vj) and is equal t
i
(L. P. C&p.2)=D 2 (l+C2C2R P 2 *)=D 2 E V2 , .... (16
and for the third stage (F/ Vj) or,
j_
(L. P. Cap.3)=3(l+C3C3# P 3 7 )=>3# P 3. .... (16;
The volumetric efficiency of 1st stage is
Volumetric efficiency of second stage is
Volumetric efficiency of third stage is
The work of the threestage compressor with the assistance of Eq. (163) ma
be stated in terms of supply pressure, pounds per square inch, displacemen
of firststage cylinder, in cubic feet and volumetric efficiency of first stage, am
also ratios of compression existing in the first, second, and third stages,
. (16?.
To make use of this formula for the work of the compressor the tw
receiver pressures must be known, and it is, therefore, important to derive j
relation between receiver pressures, displacements and clearances or volumetr:
efficiencies.
The assumption of perfect intercooling which has already been made U
of in obtaining Eq. (169), regardless of the receiverpressure, requires that
seeEq. (157):
(L.P. Cap.i) (sup.pr.) = (L.P. Cap. 2 ) (1 rec.pr.)
= (L.P. Cap. 3 )(2rec.pr.). (170
WOEK OF COMPRESSORS
137
Using values of capacities in Eqs. (163), (164), and (165) and solving for
urst and secondreceiver pressures.
nd
.
(2 rec.pr.)
(L. P. Cap.i)
(L. P. Cap.i)
.
R p i =
(1 rec.pr.) _D\E v i
(sup.pr.) ~D 2 E v2 '
(2 rec.pr.) D 2 E v2
p (1 rec.pr.)
>y definition, (del.pr.) =R P (sup.pr.),
r, (171)
D ^. (172)
(173)
(174)
(del.pr. _ (sup.pr.) _
"2>3 /o \ ** / P /r \ "i
(2 rec.pr.) (2 rec.pr.)
. . . (175)
'he work of the threestage compressor may then be stated in terms of
supply pressure, pounds per square inch, displacements, cubic feet, volumetric
fficiencies, and overall ratio of compression, R P} as follows :
sl
In Terms of Pressures, Displacements, and Clearances, an expression can be
'ivritten by substitution of values of E v \, E v2 and E v % from Eqs. (166), (167) and
J168), but it becomes a long expression, further complicated by the fact that
$ P i, R p2 and ^3 remain in it. This may be solved by the approximation
ipased first upon the assumption that all volumetric efficiencies are equal to each
!>ther or to unity when
#1
D 2
&2 (If volumetric efficiencies are each equal to each other
D 3 or to unity) (177)
138 ENGINEERING THERMODYNAMICS
This process amounts to the same thing as evaluating E v i, E V 2, and E v z fi
Eqs. (166), (167) and (168), making use of the approximation Eq. (177) anc
substituting the values found in Eq. (176).
Since the above can be done with any expression which is in terms of volu
metric efficiencies, the following formulae will be derived from Eq. (176), as i
stands.
The mean effective pressure of the threestage compressor reduced to the first
stage cylinder is found by dividing the work of the entire cycle, Eq. (176) Ir
displacement of the first stage, and by 144 to reduce to pounds per square inch
(m.e.p.) reduced to first stage cylinder,
W s
Note here that this may also be obtained by multiplying (work per cubic foe
supplied) by (volumetric efficiency of first stage) and dividing the product by 14^
The indicated horsepower of a compressor performing n cycles per minul
will be equal to the work per cycle multiplied by n and divided by 33,000, o
for the threestage compressor with general receiver pressures,
o VP ^. F ' 1 /ntf\.i
LH.P. = 2^
For n may be substituted the number of revolutions per minute, N, divid<
by the revolutions required to complete one cycle
N
.
z
The horsepower per cubic foot of gas suppliedjper minute is
LH.P. _s _ (sup.pr.) [ (DiE,i\ ^ , (D^} ^
n(L.P.Cap.) s1 229.2 l\ ' '
Horsepower per cubic foot gas delivered and cooled per minute is
I.H.P. s (del.pr.) \/DiE v i\^ L /D 2 E V ,
n(H.P.Cap.cold) s1 229.2 \\D 2 E v2 ) r \DMt t
81
WORK OF COMPRESSORS
lorse power per cubic foot hot gas delivered per minute is
139
I.H.P.
_ s (sup.pr.)
nCH.P.Cap.hot)**! 229.2
81
"*
L ["/>!#, A
* _ \BJt*J
:Fhe last equation is obtained by means of the relation
(L. P. Cap.) = (H. P. Cap. hot) X
Vint^ V
. not; X
X
.pr. \ sup.pr
(183)
// clearance is zero or negligible, these expressions may be rewritten, putting
&>, ^2 and E'vs each equal to unity.
B
H.P. per cubic foot of gas supplied per minute is
n(L.P.Cap.) 229.2
H.P. per cubic foot delivered and cooled per minute is
, .
I.H.P. _ (del, pr.) [7>i
sl
sl
pr.)r/J) 1
.2 [\5i
n(H.P.Cap.cold)~ 229
3. P. per cubic foot hot gas delivered per minute is
I.H.P. g (sup.pr.) /DA ^ if/ftX" , /
n(H.P. Cap.hot) sl 229.2 \Z) 3 / "l\ft
sl
8T]
140
ENGINEERING THERMODYNAMICS
Example 1. Method of calculating Diagram, Fig. 35.
Assumed data:
P a = Pb =2116 Ibs. per square foot.
P c =p d =P l =P m =4330 Ibs. per square foot.
P e =P f = Pj=P t =S830 Ibs. per square foot.
P g =p h = 18,000 Ibs. per square foot.
d=7.5 per cent for all cylinders; s = 1.4.
L.P. capacity 5 cu.ft.
To obtain point M:
From formula Eq. (163) L. P. Capi. = A(l +Ci CiR p i )
5 =Di(l +.075 .075x1.67) or Z>i= 5.3 cu.ft. and clearance volume
F m =5.3x.075=.387cu.ft.
or
V m = .39 cu.ft. ; P m =4330 Ibs. sq.ft. ;
~ i 4 . =.39X1.67 = .67 cu.ft.
Therefore,
To obtain Point A :
Additional points M to A may be found by assuming pressures and finding cori
spending volumes as for F a .
To obtain point B :
Therefore,
To obtain point C:
. P. Capi.) =.67+5 =5.67 cu.ft.
F 6 =5.67 cu.ft.; P b =2116 Ibs. sq. ft.;
 =5.6751.67 =3.45 cu.ft.
Therefore,
V c = 3.45 cu.ft. ; P c = 4330 Ibs. sq.ft.
Intermediate points B to C may be found by assuming various pressures and find]
corresponding volumes as for F&.
WORK OF COMPRESSORS 141*
obtain point D :
Volume at D is the displacement plus clearance of the intermediate cylinder. This
cmot be found until the capacity is known. Applying the same sort of relations as
\? re used in calculating the diagram for the twostage case with clearance,
1
D 2 (l+C2C 2 Rp2) =2.44 or D 2 =2.57,
id clearance volume.
V k = . 075X2.57. 192 cu.ft.,
tnce,
V d =2.57 +.19 =2.76 cu.ft.
lierefore,
V d =2.76 cu.ft; P d =4330 Ibs. sq.ft.;
The rest of the points are determined by methods that require no further explana
t<n and as pressures were fixed only volumes are to be found. These have the following
^lues, which should be checked:
1=1.65; 7, = 1.32; 7, = .79; 7*=.09; 7, = .15; 7, = .32; 7* = .65; 7,, = 1.23; V z = . 14.
Example 2. A threestage compressor is compressing air from atmosphere to 140
k per square inch absolute. The lowpressure cylinder is 32x24 ins. and is known to
Ive a clearance of 5 per cent. From gages on the machine it is noted that the first
Reiver pressure is 15 Ibs. per square inch gage and the secondreceiver pressure is
i Ibs. per square inch gage. What horsepower is being developed if the speed is
R.P'.M. and s = 1.4? From the formula Eq. (169),
9 r 81 SjJ. sl
W = 144(sup.pr.)D 1 # Pl [# pl +R P2 s +R PS ,
1 From gage readings
7? 3 2 7? 7 2^ 7? 14 2
^ = ^=2. ^ 2=3o= 2.33, R P3  2.
i
^ n (l+CiCifti7) fromEq. (166),
E n = (l +.05 .05X1. 65) =67.5 per cent,
f 'ence,
W = 144x^ Xl5xll.2x.675(1.22+1.28 + 1.223);
= 59,200 ft.lbs. per stroke or 200 X59,200 ft. = Ibs. per minute;
= 358 I.H.P.
142 ENGINEERING THERMODYNAMICS
Examples. Another compressor has cylinders 12x20x32x24 in. and it is know
that the volumetric efficiencies of the high, intermediate and lowpressure cylinders ai
respectively 70 per cent, 85 per cent and 98 per cent. The (del.pr.) is 150 Ibs. per squar
inch absolute. What is the horsepower in this case if the speed is 100 R.P.M.?
From the formula Eq. (176),
= (1.309+1.495+1 3) =66,400 ft.lbs. per stroke,
200X66,400
Whence I.H.P. = 33^5 =402.
Prob. 1. What will be the horsepower required to drive a 12 X22 X34 X30 in. th
stage compressor with volumetric efficiencies of 75, 85, and 95 per cent in the high
intermediate and lowpressure cylinders, at 100 R.P.M. when compressing natural ga
from 25 Ibs. per square inch gage to 300 Ibs. per square inch gage, adiabatically?
Prob. 2. A threestage compressor for supplying air for a compressedair locomc
tive receives air at atmosphere and delivers it at 800 Ibs. per square inch gage. Shoul
the receiver pressures be 50 Ibs. and 220 Ibs. respectively in the first and second and th
volumetric efficiency of the first stage 90 per cent, what would be its displacemei}
and horsepower when compressing 125 cu.ft. of free air per minute, adiabatically
What are the cylinder displacements?
Prob. 3. Find the work done on a gas, the value for s of which is 1.3, in compressiD
it from atmosphere to 7 atmospheres absolute, adiabatically in three stages, the com
pressor having a lowpressure cylinder displacement of 60 cu.ft. per minute and a voli
metric efficiency of 95 per cent, first receiver pressure being 2 atmospheres absolut*
and secondreceiver pressure 4 atmospheres absolute. If air were being compresse
instead of the above gas, how would the work vary?
Prob. 4. The cylinders of a compressor are 8x12x18x24 ins. and clearance suo
as to give volumetric efficiencies of 80, 90 and 98 per cent in the different cylinders in tl:
order given. Compressor is double acting, running at 120 R.P.M. and compressing ai
adiabatically from 14 Ibs. per square inch absolute to 150 Ibs. per square inch gag(
What is the capacity in cubic feet per minute, work done per cubic feet of (sup.pr.) aii
(del.pr.) air hot and cold and the horsepower of the compressor? .What would t
the effect on these quantities if the clearances were neglected?
Prob. 5. If the cylinders of a compressor are 10x14x20x18 ins., and clearanc*
are 8, 5 and 3 per cent, what work is being done in adiabatically compressing air fro:
10 Ibs. per square inch absolute to 100 bs. per square inch gage?
NOTE: Solve by approximate method.
Prob. 6. For special reasons it is planned to keep the firstreceiver pressure of
threestage compressor at 30 Ibs. per square inch absolute, the secondreceiver pressure :
60 Ibs. per square inch absolute, and the line pressure at 120 Ibs. per square inch absolu
WOEK OF COMPRESSORS 143
le (sup.pr.) is 14 Ibs. per square inch absolute. If the clearances are 4 per cent in
t: low and 8 per cent in the intermediate and highpressure cylinders, what must be
t ! cylinder sizes to handle 500 cu.ft. of free air per minute at 120 R.P.M. and what
p ^er must be supplied to the compressor on a basis of 80 per cent mechanical effi
cacy, for a value of s equal to 1.39?
Prob. 7. Should the above pressures (Prob. 6) be gage pressures instead of absolute,
r N would the quantities to be found be affected?
Prob. 8. The receiver pressures on a C0 2 gas compressor are 50 Ibs. per square inch
r solute, and 200 Ibs per square inch absolute, the (del.pr.) being 1000 Ibs. per square
i h absolute. The mach ne has a lowpressure cylinder 8x10 ins. with 3 per cent
carance. What horsepower will be required to run it at 100 R.P.M. and what would
1 the resultant horsepower and capacity if each pressure were halved? (Sup.pr.) = 14.7
1 . per square inch.
16. Threestage Compressor with Bestreceiver Pressures Exponential
(mpression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure
ad Horsepower in Terms of Dimensions of Cylinders and Clearances. It
vs found that for the threestage adiabatic compressor with perfect inter
CDling, the work was a minimum if the first and second receivers had pressures
cfined as follows, see Eqs. (147) and (148) :
(best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*. ..... (188)
(best 2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]*. ..... (189)
, _ (best 1 rec.pr.) _ / del.pr.X* _
pl ~ (sup.pr.)
(best 2 rec.pr.) /del.prA*
2 = 7i i = K>J> ....
(best 1 rec.pr.) \sup.pr./
(del.pr.) /deLprA*
= 7r~^^ \ = \ I =
(best 2 rec.pr.) V sup.pr./
The use of these values in connection with expressions previously given
f< volumetric efficiency, Eqs. (166), (167) and (168), gives,
\lumetricefficiencyoffirststage =E c i = (l+c 1 c 1 R/~*) .... (193)
Ylumetric efficiency of second stage = #2 = (1+ C 2 C 2#? 3s ) .... (149)
V.lumetric efficiency of third stage =^ P 3 = (l+c 3 c 3 R P 3s ) .... (195)
144
ENGINEERING THERMODYNAMICS
The work of the threestage compressor with bestreceiver pressures, E(
(162), when expressed in terms of displacement and volumetric efficiency beconu
s 1
s
where
(del.pr.)
7 ^
(sup.pr.)
If clearance is known, the value of E,\ may be ascertained by Eq. (19;
and inserted in Eq. (196). Since this may be so readily done the substitutic
will not here be made.
The mean effective pressure of the compressor referred to the first stage
obtained by dividing the work Eq. (196) by 144 DI:
(m.e.p.) referred to firststage cylinder
W
1
(19
The mean effective pressures of the respective stages, due to the equali
of work done in the three stages will be as follows:
For first stage
(m.e.p.) = (sup.pr.)^! (fl/STl)
o 1
(K
For second stage
(m.e.p.) =
 1)
For third stage
(m.e.p.) = ^ (sup.pr.)^.i (72* V 1).
But
.prO'* = (1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*,
and also
.pr.) = (2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]*.
WORK OF COMPRESSORS
lence
'or second stage
s1
(m.e.p.) =:[(sup.pr.) 2 (deLpr.)]^2(^ 1).
i.
third stage
145
(201)
(202)
Conditions to Give Bestreceiver Pressures. All the foregoing discussion
;>f bestreceiver pressures for the threestage compressor can apply only to cases
;Q which all the conditions are fulfilled necessary to the existence of bestreceiver
oressures. These conditions are expressed by equations (173), (174), (175),
190), (191), and (192), which may be combined as follows:
(1) (2) (3) (4)
(L. P. Cap.i; = (L. P. Cap. 2 ) = DiE vl =
(L.P.Cap.2) (L.P.Cap.3)
(5)
(6)
. . (203)
'arts (1) and (2) of this equation state the requirements in terms of
capacities; (3) and (4) in terms of displacements and volumetric efficiencies;
.'5) and (6) in terms of displacements and clearances. In order, then, that best
Deceiver pressure may be obtained, there must be a certain relation between
:he given ratio of compression and dimensions of cylinders and clearances.
Since, after the compressor is once built these dimensions are fixed, a given
nultistage compressor can be made to give bestreceiver pressures only when
compressing through a given range, i.e., when R p has a definite value. If R p
has any other value the receiver pressures are not best, and the methods of the
previous Section (15) must be applied.
When clearance percentages are equal in all three cylinders, ci=C2=cs, and
the volumetric efficiencies are all equal then, when bestreceiver pressures exist,
Eq. (203) becomes,
RJ j^ = j^ = for equal clearance per cent. . . (204)
Evidently this same expression holds if clearances are all zero or negligible.
What constitutes negligible clearance is a question requiring careful thought
and is dependent upon the ratio of compression and the percentage of error
allowable.
146 ENGINEERING THERMODYNAMICS
Indicated horsepower of the compressor is found by multiplying the wor
per cycle, Eq. (196) by the number of cycles per minute, n, and dividing th
product by 33,000.
l) .... (20fi
From this are obtained the following:
H.P. per cubic foot supplied per minute
I.H.P. s (sup.pr.) , ii^ 1
H.P. per cubic foot delivered and cooled per minute
I.H.P. s (del.pr.) fzi 1
n(H. P. Cap. cold) = 7^l ~J&T (Rp * 1} ..... (20 '
H.P. per cubic foot delivered hot per minute
(sup.pr.) 2 il =!
=: K v 3s (H p 3s 1). . . (ZOl
(See Eq. (156)).
It is useful to note that these expressions are all independent of clearance
which is to be expected, since the multistage compressor may be regarded as
series of singlestage compressors, and in single stage such an independenc
was found for work and horsepower per unit of capacity.
Example. If the following threestage compressor be run at bestreceiver pressure
what will be the horsepower and the bestreceiver pressures? Compressor has lov
pressure cylinder 32x24 ins. with 5 per cent clearance, is compressing air from atmo>
phere to 140 Ibs. pier square inch absolute, so that s equals 1.4 and it runs at 100 R.P.&
From the formula Eq.~(196)
From the formula Eq. (188)
(best 1 rec.pr.)= [(sup.pr.) 2 (del.pr.)]i
= (15) 2 X140]*=31.6.
WORK OF COMPRESSORS 147
From Eq. (189)
(best 2 rec.pr.) =[ (sup.pr.) (del.pr.) 2 ]*
= [15X(140) 2 ]4=66.5.
From Eq. (193)
i
;mco,
W =432 X^ X15 xll.2 X96.5 X (9.35 95 l) =59,000 ft.lbs.,
, p 59,000X200
33,000
Prob. 1. There is available for running a compressor 175 H.P. How many cu.ft.
' free air per minute can be compressed from atmosphere to 150 Ibs. per square inch
ige by a threestage adiabatic compressor with bestreceiver pressures?
Prob. 2. The lowpressure cylinder of a threestage compressor has a capacity of
i cu.ft. per stroke. If the stroke of all three cylinders is 18 ins., what must be the
ameters of the intermediate and high to insure bestreceiver pressures, if clearance
.5 neglected, and (sup.pr.) be 1 Ib.jper square inch absolute and (del.pr.) 15 Ibs. per
luare inch absolute, s being 1.4.
Prob. 3. The above compressor is used as a dryvacuum pump for use with a sur
ce condenser. If 800 cu.ft. of (sup.pr.) gas must be handled per minute what horse
>wer will be needed to run it? What will be the horsepower per cubic foot of atmos
jieric air?
Prob. 4. Will a 15 X22 X34 X24 in. compressor with clearances of 3, 5 and 8 per cent
low, intermediate and highpressure cylinders respectively be working at bestreceiver
ressures when (sup.pr.) is 15 Ibs. per square inch absolute and (del.pr.) 150 Ibs. per
jiuare inch absolute? If not, find by trial, the approximate (del.pr.), for which this
achine is best, with s equal to 1.4?
Prob. 5. For the best (del.pr.) as found above find the horsepower to run the
achine at 100 R.P.M. and also the horsepower per cu.ft. of (del. pr.) air cold?
Prob. 6. Should this compressor be used for compressing ammon'a would tl.c
')st (del. pr.) change, and if so what would be its value? Also what power would Le
;eded for this case?
Prob. 7. Compare the work necessary to compress adiabatically in three stages from
') Ibs. per square inch absolute to 200 Ibs. per square inch absolute, the following gases:
Air; Oxygen; Gasengine mixtures, for which s = 1.36.
Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio
:r bestreceiver pressure and a pressure ratio of 10?
148 ENGINEEEING THERMODYNAMICS
Prob. 9. A compressor, the lowpressure cylinder of which is 30x20 ins. with 5 pi
cent clearance is compressing air adiabatically from atmosphere to 150 Ibs. per squa:
inch gage, at bestreceiver pressure. Due to a sudden demand for air the (del. pr
drops to 100 Ibs. per square inch gage. Assuming that the (1 rec. pr.) dropped to 5 lb
per square inch gage and (2 rec.pr.) dropped to 40 Ibs. per square inch gage, how muc
would the speed rise if the power supplied to machine was not changed?
17. Comparative Economy or Efficiency of Compressors. As the prin
duty of compressors of all sorts is to move gas or vapor from a region of IQI
to a region of high pressure, and as this process always requires the expenditui
of work, the compressor process which is most economical is the one thi
accomplishes the desired transference with the least work. In this sense, thei
economy of compression means something different than efficiency, as ord
narily considered. Ordinarily, efficiency is the ratio of the energy at one poii
in a train of transmission or transformation, to the energy at another poin
whereas with compressors, economy of compression is understood to mean tl:
ratio of the work required to compress and deliver a unit of gas, moving
from a low ,to a highpressure place, to the work that would have been require
by some other process or hypothesis, referred to as a standard. This econom
of compression must not be confused with efficiency of compressors as machine
as it is merely a comparison of the work in the compressor cylinder for an actu;
case or hypothesis to that for some other hypothesis taken as a standard. Tl
standard of comparison may be any one of several possible, and unfortunate!
there is no accepted practice with regard to this standard. It will, therefor
be necessary to specify the standard of reference whenever economy of compre
sion is under consideration. The following standards have been used wit
some propriety and each is as useful, as it supplies the sort of information reall
desired.
First Standard. The work per cubic foot of supply gas necessary to COD
press isothermally (Cycle 1), from the supply pressure to the delivery pressu
of the existing compressor and to deliver at the high pressure is less than that
any commercial process of compression, and may be taken as a standard f
comparison. Since, however, actual compressors never depart greatly fro*
the adiabatic law, their economy compared with the isothermal standard w/
always be low, making their performance seem poor, whereas they may be '
nearly perfect as is possible, so that it may appear that some other standa
would be a better indication of their excellence.
Second Standard. The work per cubic foot of gas supplied when compresst
adiabatically in a single stage (Cycle 3), if taken as a standard, will indicate
high economy, near unity for singlestage compressors, and an economy abo
unity for most multistage compressors. For the purpose of comparison
will be equally as good as the first standard, and the excess of the economy ov
unity will be a measure of the saving over singlestage adiabatic compressic
Since, however, singlestage adiabatic compression is not the most economic
obtainable in practice for many cases, this standard may give an incorrc
idea of the perfection of the compressor.
WORK OF COMPRESSORS 149
Third Standard. Due to the facts noted above, it may be a better indica
on of the degree of perfection of the compressor to compare the work per
ibic foot of gas supplied with that computed for the standard adiabatic cycle
ost nearly approaching that of the compressor. This standard is, however,
)en to the objection that a multistage compressor is not referred to the same
ycle as a singlestage compressor, and a multistage compressor with other
lan bestreceiver pressure is not referred to the same cycle as another operating
; ith bestreceiver pressure. This is, therefore, not a desirable standard for
omparing compressors of different types with one another, although it doe s
low to what extent the compressor approaches the hypothetical best condi
,on for its own type and size.
Other standards might be chosen for special reasons, each having a value
i proportion as it supplies the information that is sought.
It is seen from the discussion of the second standard that its only advantage
;ver the first is in that it affords a measure of the saving or loss as compared
4th the singlestage adiabatic compressor cycle.
If the first standard, that of the isothermal compressor cycle, be adopted
3r the purpose of comparison, it at once gives a measure of comparison with
he isothermal, which is more "and more nearly approached as the number
f stages is increased, though never quite reached, or as the gas is more effect
yely cooled during compression. It may be regarded as the limiting case of
lultistage compression with perfect intercooling, or the limiting case of con
inuous cooling. ,
In order to ascertain how nearly the actual compressor approaches the
Adiabatic cycle most nearly representing its working conditions, the economy of
f the various reference cycles heretofore discussed may be tabulated or charted,
,nd the economy of the cycle as compared with that of the actual performance
>f the compressor will give the required information. The process of com
mtation by which this information is obtained will depend upon the nature
j)f information sought. The economy of actual compressor compared with the
isothermal may be stated in any of the following ways:
Computed work per cubic foot supplied, isothermal ( .
Indicated work per cu.ft. actual gas supplied to compressor
I.H.P. per cubic foot per minute supplied, isothermal
I.H.P. per cubic foot per minute actual supplied
Single stage
(209)
(m.e.p.) isothermal, pounds per square inch, no clearance
. _ _ ( {*}
(m.e.p.) actual r true volumetric efficiency
Multistage
(m.e.p.) isothermal, no clearance
(m.e.p.) reduced to first stage 4 first stage vol. eff.
150 ENGINEERING THERMODYNAMICS
In this connection it is useful to note that for the case of the noclearan
cycles, the work per cubic foot of supply is equal to the mean effective pressu
(M.E.P.) in pounds per square foot, and when divided by 144 gives (m.e.j
in pounds per square inch. Also, that in cases with clearance, or even actu
compressors with negligible clearance, but in which, due to leakage and oth
causes, the true volumetric efficiency is not equal to unity,
Work per cubic foot gas supplied X E v =1 44 (m.e. p.). . . (21
The information that is ordinarily available to determine the econon
of the compressor will be in the form of indicator cards from which the (m.e.f
for the individual cylinders may be obtained with ordinary accuracy. TJ
volumetric efficiency may be approximated from the indicator cards also, b
with certain errors due to leakage and heating, that will be discussi
later. If by this or other more accurate means the true volumetr
efficiency is found, the information required for the use of Eq. (209) (
or (d) is available. Evaluation of the numerator may be had by Eq. (31
which is repeated below, or by reference to the curve sheets found at the ei
of this chapter. (Fig. 50.)
Mean effective pressure, in pounds per square inch for the isothermal coi
pressor without clearance is given by
(m.e.p.) isothermal =(sup.pr.) log e R P (21
The curve sheet mentioned above also gives the economy of adiabat
cycles of single stage, also two and three stages with bestreceiver pressure
The value of s will depend upon the substance compressed and its conditio
The curve sheet is arranged to give the choice of the proper value of s applyii
to the specific problem.
If it is required to find the economy of an actual compressor referred
the third standard, i.e., that hypothetical adiabatic cycle which most near
approaches the actual, then
Economy by third standard is
Econ. actual referred to isothermal
Econ. hypothetical referred to isothermal*
. . (21
It is important to notice that for a vapor an isothermal process is not o<j
following the law PXV = constant. What has, in this section, been called
isothermal is correctly so called only so long as the substance is a gas. Sini:
however, the pressurevolume analysis is not adequate for the treatment
vapors, and as they will be discussed under the subject of Heat and Woi
Chapter VI, it is best to regard this section as referring only to the treatmc
of gases, or superheated vapors which act very nearly as gases. Howev
it must be understood that whenever the curve follows the law PXF = consta:
the isothermal equations for work apply, even if the substance be a vaf
and the process is not isothermal.
WORK OF COMPRESSORS 151
18. Conditions of Maximum Work of Compressors. Certain types of com
ressors are intended to operate with a delivery pressure approximately con
tant, but may have a varying supply pressure. Such a case is found in pumps
r compressors intended to create or maintain a vacuum and in pumping
tatural gas from wells to pipe lines. The former deliver to atmosphere, thus
laving a substantially constant delivery pressure. The supply pressure,
lowever, is variable, depending upon the vacuum maintained. In order that
uch a compressor may have supplied to it a sufficient amount of power to
;eep it running under all conditions, it is desirable to learn in what way this
>ower required will vary, and if it reaches a maximum what is its value, and
,inder what conditions.
Examine first the expression for work of a singlestage adiabatic compressor
yith clearance. The work per cycle will vary directly as the mean effective
Pressure. Eq. (69.)
(213)
sup.pr. sup.pr
This will have a maximum value when
^(sup.pr.)
>r when
/deLprA^ , f 1+c _^/deLprAri
\sup.pr./ 1fc L s \sup.pr./ J
Solving this for the value of supply pressure will give that supply pressure
it which the work will be a maximum, in terms of a given delivery pressure,
clearance and the exponent s.
The assumption most commonly used is that clearance is zero. If this is
rue~or the assumption permissible, the above equation becomes simplified,
(sup.pr.
(215)
Che value of s for air, for instance, is 1.406, and hence the ratio of compression
i'or maximum work for the hypothetical air compressor is
3 ' 46
(1.406)' = 3.26 ......... . (216)
It may be noted that when s = 1 in the above expression, the value of the
;atio of compression become indeterminate. To find the supply pressure for
naximum work in this case, take the expression for mean effective pressure
'or the isothermal compressor (s = l), Eq. (43),
. . . ( 2 17)
152 ENGINEEKING THERMODYNAMICS
Differentiate with respect to (sup.pr.) and place the differential coefficient equa
to zero. This process results in the expression
loge  + _ =
' Vsup.pr./ 1fc Vsup.pr.
When c=0, this becomes,
/deLpr.X l or
Vsup.pr./
The expressions Eqs. (215) and (219) given are easily solved, but Eqs. (214
and (218) are not, and to facilitate computations requiring their solution
the results of the computation are given graphically on the chart, Fig. 48, a
the end of this chapter.
The mean effective pressure for a compressor operating under maximun
work conditions may be found by substituting the proper ratio of compressior
found as above, in Eq. (213) or (217). In Fig. 48 are found also the result
of this computation in the form of curves. Note in these curves that the mea:
effective pressure is expressed as a decimal fraction of the delivery pressure
The discussion so far applies to only singlestage compressors. The probler
of maximum work for multistage compressors is somewhat different, and it
solution is not so frequently required. Moreover, if the assumption of perfec
intercooling is made, the results are not of great value, as a still greater amoun
of power might be required, due to the failure for a period of time of the suppl;
of cooling water. Consider this case first.
If intercooling be discontinued in a multistage compressor, the volum
entering the second stage will equal that delivered from the first, and similar!
for the third and second stages. The entire work done in all stages will be thi
same as if it had all been done in a single stage. It might be questioned a
to whether this would hold, when the ratio of compression is much less tha:i
designed. The first stage will compress until the volume has become as sma;
as the lowpressure capacity of the second stage. If the delivery pressure i^
reached before this volume is reached, there is no work left to be done in th
second or any subsequent stages, and, due to the pressure of the gas, the valves
if automatic, will be lifted in the second and higher stages, and the gas m}
be blown through, with only friction work. It appears then that under till
condition of no intercooling the multistage compressor acts the same as i
single stage, and the conditions of maximum work will be the same.
If intercooling is maintained perfect there will still be a range of pressure
on which all the work of compression is done in the first stage, merely blowing
the discharge through receivers, valves and cylinders in the upper stages. 1
this range is such that this continues beyond a ratio of pressures, which gives
maximum (m.e.p.) for the single stage, then the maximum will have been reachej
while the compressor is operating single stage, and the singlestage fonnuk
and curves may be applied to this case also.
WOEK OF COMPRESSORS
153
That this condition frequently exists with multistage compressors of
c linary design is shown by the fact that the ratio of compression in each stage
i seldom less than 3, and more frequently 3.5, 4 or even more. The ratio
c compression giving maximum work for single stage, has values from 2.5
t 3.26, dependent on clearance and the value of 6' for the gas compressed, and
i therefore, less for the majority of cases.
, If a curve be drawn, Fig. 36, with ratio of compression as abscissas and
(i.e.p.^del.pr.) as ordinates, so long as the action is single stage, a smooth curve
vll result, but when the ratio of compression is reached above which the second
(Under begins to act, the curve changes direction suddenly, falling as the ratio
A2
.40
.38
7
Max for Two Stage Cpmpressor
Cylinder Ratic
Single
Stage
Values of R P .
36. Curve of Relation between Mean Effective Pressure per Pound of Delivery Pressure
in Terms of Pressure Ratio for Air, showing Maximum Value.
Decompression increases. Hence, if the ratio of cylinders is such that the single
* go maximum is not reached before the second stage begins to operate, the highest
pint of the curve, or maximum work for a given delivery pressure will occur
v>en the ratio of supply and delivery pressures is such as to make firstreceiver
p'ssure equal to delivery pressure.
19. Actual Compressor Characteristics. Air or gas compressors are very
cnmonly made double acting, so that for a single cylinder, two cycles will
b performed during one revolution, one in each end of the cylinder. If a rod
e;ends through one of these spaces and not through the other, the displace
n nt of that end of the cylinder will be less than the other by a volume equal
tt. the area of rod multiplied by the stroke. To avoid mechanical shock
I'the end of either stroke, it is necessary to leave some space between the
154 ENGINEERING THERMODYNAMICS
piston and cylinder head. Passages must also be provided, communicahj
with inlet and discharge valves. The total volume remaining in this sp
and in the passages when the piston is at the nearer end of its stroke constiti^
the clearance. The amount of this clearance volume varies from .5 or .($
one per cent in some very large compressors to as much as 4 or 5 per ceni
the volume of displacement in good small cylinders.
In order to study the performance of an actual compressor and to comj
it with the hypothetical cycle, it is necessary to obtain an indicator card, :i
knowing the clearance and barometric pressure to convert the indicator C'<
into a pressure volume diagram, by methods explained in Chapter I. Fig. 7
is such a diagram for a singlestage compressor. In the pipe leading to .1
H.P.Cap Hot (Apparent)
(Sup. Pr.)
Vol.
FIG. 37. Compressor Indicator Card Illustrating Departure from Reference Cycle.
intake valve the pressure is determined and a horizontal line AB is dr/i
on the diagram at a height to represent the supply pressure. Simik v
discharge pressure is determined and drawn on the diagram, KE. Cons ;
the four phases of this diagram in succession.
1. Intake Line. At a point somewhat below A the intake valve op is
say at the point F. This remains open till a point H is reached at or near
end of the stroke. The line connecting these two points indicates varial iv
of pressures and volumes throughout the supply stroke. In general this
will lie below the supplypressure line AB due to first, the pressure necesft
to lift the inlet valve from its seat against its spring and inertia, if autom
and support it, and second, friction in the passages leading to the cylinder 1
the point where the supply pressure was measured. While the former is m b
constant the latter varies, depending upon the velocity of gases in the pass* *
The piston attains its highest velocity near the middle of the stroke, u:
WORK OF COMPRESSORS 155
(using the intake line to drop below the supply pressure more at this part
c stroke. These considerations do not, however, account completely for the
irm of the intake line. Frequently the first portion of the line lies lower than
1e last portion, even at points where the piston velocities are equal. This
i more prominent on a compressor having a long supply pipe, and is due to
is forces required to accelerate the air in the supply pipe while piston velocity
i increasing, and to retard it while piston velocity is decreasing. In com
jessors where the inlet valve is mechanically operated and the supply pipe
hg, it is possible to obtain a pressure at the end of the intake, line H, even
i excess of the supply pressure. The effect of this upon volumetric efficiency
\11 be noted later.
The apparent fluctuations in pressure during the first part of the intake
1 e may be attributed, first, to inertia vibrations of the indicator arm, in which
.se the fluctations may not indicate real variations of pressure; second, the
iiicator card may show true variations of pressure due to inertia of the gases
i the supply pipe, since a moment before the valve opened at F the gases were
sitionary in the supply pipe. When F is reached the piston is already in motion
sd a very considerable velocity is demanded in the supply pipe to supply
t.3 demand. This sudden acceleration can be caused only by a difference in
pssure, which is seen to exist below and to the right of F on the diagram,
ne suddenness of this acceleration may start a surging action which will cause
re and fall of pressure to a decreasing, extent immediately after. A third cause
i possible, that is, a vibration of the inlet valve due to its sudden opening
\ien it is of the common form, mechanical valves change the conditions. It
iclosed by weight or a spring and opened by the pressure difference. Between
tese forces the valve disk may vibrate, so affecting the pressure.
2. Compression Line. From the time the inlet valve closes at the point
/ until the discharge valve opens at the point G, the gases within the cylinder
ai being compressed. The compression is very nearly adiabatic in ordinary
jactice, but due to the exchange of heat between the cylinder walls, at first
fun walls and later from gas to walls, which are cooled by water jacket to
Invent the metal from overheating, there is a slight departure from the adia
1 tic law almost too small to measure.
> j A second factor which influences the form of this curve to a greater extent
Ueakage. This may occur around the piston, permitting gas to escape from
qe end of the cylinder to the other. During the compression process there is
fut an excess of pressure in the other end of the cylinder due to reexpansion,
tiding to increase pressures on the first part of compression. Later, the
pssure rises and the pressure on the other side of the piston falls to supply
p3ssure. During this period leakage past piston tends to decrease successive
Fissures or lowers the compression line. Leakage also occurs through either
c;charge or inlet valves. The former will raise the compression line, while
e3essive leakage of the inlet valve will lower it.
It is then evident that unless the nature of the leakage is known, it is
[(possible to predict the way in which it will change this line. It is, however,
!56 ENGINEERING THERMODYNAMICS
WORK OF COMPRESSORS
157
[valve. After reexpansion is completed the intake valve opens and gas enters
the end of the cylinder under consideration. At the same time compression
I is taking place in the other end, and later deliver}'. During these processes
j whatever gas leaks past the piston tends to fill the end of the cylinder in which
intake is going on. Leakage past the discharge valve also tends to fill the cylin
der with leakage gas. Both of these tendencies decrease the quantity of gas
! entering through this intake valve, and its true amount when reduced to external
; supply pressure and temperature is, therefore, less than the volume AB.
The true lowpressure capacity of the compressor is the true volume of gas
I under external supply conditions that enters the cylinder for each cycle. This
i cannot be determined from the indicator card except by making certain assump
tions which involve some error at best. It can, however, be ascertained by
means of additional apparatus, such as meters or calibrated nozzles or receivers,
by means of which the true amount of gas compressed per unit of time is made
known. This reduced to the volume per cycle under supply pressure and tem
perature will give the true lowpressure capacity.
Volumetric efficiency is defined as being the ratio of lowpressure capacity
! to displacement. On the diagram, Fig. 37, the displacement is represented
Ito the volume scale by the horizontal distance between verticals through the
extreme ends of the diagram, K and H. Since there are three ways in which
the lowpressure capacity may be approximated or determined, there is a
corresponding number of expressions for volumetric efficiency.
1. The volumetric efficiency of the hypothetical cycle is
^(hypothetical) =
(hypothetical L. P. Cap.)
(displacement)
. . (220)
and this is evaluated and used in computations in the foregoing sections of
I this chapter.
2. The apparent volumetric efficiency is
(apparent L. P. Cap.) / 991 x
=  (22.
, N
R(apparent)
and would be very nearly equal to the true volumetric efficiency were it not
for leakage valve resistance and heating during suction, but due to this may
be very different from it.
3. The true volumetric efficiency is
(true L. P. Cap.)
(displacement ) '
(222)
In problems of design or prediction it is necessary either to find dimensions,
F( speeds and power necessary to give certain actual results, or with given dimer
* ions and speeds to ascertain the probable power and capacity or
15 s ENGINEERING THERMODYNAMICS
characteristics of actual performance. Since it is impossible to obtain actual
performance identical with the hypothetical, and since the former cannot be
computed, the most satisfactory method of estimate is to perform the computa
tions on the hypothetical cycle, as is explained in previous sections of this chap
ter, and then to apply to these results factors which have been found by
comparing actual with hypothetical performance on existing machines as nearly
like that under discussion as can be obtained. This necessitates access to data
on tests performed on compressors in which not only indicator cards are taken
and speed recorded, but also some reliable measurement of gas compressed.
The following* factors or ratios will be found of much use, and should be
evaluated whenever such data is to be had :
# r (true)_ (true L. P. Cap.)
61 ~ E, (hypothetical) (hypothetical L. P. Cap.)"
ff p (true) (true L. P. Cap.)
~ # p (apparent) " (apparent L. P. Cap.)'
true I.H.P. true m.e.p.
~~ hypothetical I.H.P. ~ hypothetical m.e.p.'
Then
true work per cu.ft. gas, supplied
hypothetical work per cu.ft. gas, supplied
I.H.P.
true 7V
true I.H.P. per cu.ft. gas supplied = (L. P. Cap.) .^^
"hypothetical I.H.P. per cu.ft. supplied"  ptif>fll IH.P.
(L.P.Cap.)
_ true m.e.p. __.__ true L. P. Cap. _es
"hypothetical m.e.p. ' hypothetical L. P. Cap. e\
This ratio can be used to convert from hypothetical work per cubic foo
gas supplied to probable true work per cubic foot.
Multistage Compressors are subject in each stage to all of the characteristic
described for single stage to a greater or less extent. Valve resistance, frictio)
and inertia affect the intake and discharge lines; heat transfer and leakag
influence the form of compression and reexpansion lines, and the true capacit
of the cylinder is made different from the apparent due to leakage, pressur
and temperatures changes.
In addition to these points it is useful to note one special way in which th
multistage compressor differs from the single stage. The discharge of the fir?
stage is not delivered to a reservoir in which the pressure is constant, but
receiver of limited capacity. The average rate at which gas is delivered t
the receiver must equal the average rate at which it passes to the next cylinde
The momentary rate of supply and removal is not constantly the same, howeve
WORK OF COMPRESSORS
159
''his causes a rise or fall of pressure. It is evident that this pressure fluctuation
3 greatest for a small receiver. Very small receivers are not, however, used
n gas compressors due to the necessity of cooling the gas as it passes from one
tage to the next. To accomplish this a large amount of cooling surface must
e exposed, requiring a large chamber in which it can be done. Thus, it is
een that the hypothetical cycles assumed for multistage compressors do not
ruly represent the actual cycle, but the difference can never be very great,
ue to the large size of receiver which must always be used.
Another way in which the performance of this multistage compressor
ommonly differs from assumptions made in the foregoing discussions is in
Del. Pr.
\c
EG. 38. Effect of Loss of Intercooling in Twostage Compressors on Receiverpressure and
Work Distribution in the Two Cylinders.
to intercooling. It seldom occurs that the gases enter all stages at
F ie same temperature. In the several stages the temperature of the gases
jjll depend on the amount of compression, on the cooling surface and on the
tount and temperature of cooling water. The effect of variations in tem
irature upon the work and receiver pressures will be taken up later. It may
noted now, however, that if all cooling water is shut off, the gas passes from
ie cylinder to the next without cooling, there is no decrease in volume in the
leiver. For simplicity take the case of zero clearance, twostage (Fig. 38).
ABCDEF be the cycle for perfect intercooling. AB and KD are the low
fressure capacities of the first and second stages respectively. If now, inter
ooling ceases, the gas will no longer change volume in the receiver. The
iver gas, in order to be made sufficiently dense to occupy the same
160 ENGINEERING THERMODYNAMICS
volume (KD) as it did before, must be subjected to a greater pressure in th<
first stage. The new receiver^ line will be K'D'. The work of the firs
stage will therefore be ~ABWK'\ of the second stage K'D'GF, and the tota
work in the new condition is greater than when intercoooling was perfect by ai
amount represented by the area DCGE.
In the case where clearance is considered, the effect is the same, except tha
the increasing receiver pressure, increasing the ratio of compression of the firs
stage, causes the volumetric efficiency of the first stage to become less, am
hence lessens the capacity of the compressor. The effect on work per unit o
capacity is the same as without clearance.
The question as to how many stages should be used for a given compresso
is dependent upon the ratio of compression largely, and so is due, first, to con
siderations of economy, which can be understood from the foregoing sections
second, for mechanical reasons, to avoid high pressures in large cylinders; third
for thermal reasons, to avoid such high temperatures that the lubrication o
the cylinders would be made difficult, or other dangers, such as explosions
involved.
Practice varies very widely as to the limiting pressures for single, tw<
three or fourstage compressors. Air compressors of a single stage are con
monly used for ratios of compression as high as 6 or 7 (75 to 90 Ibs. gage). Fc
ratios greater than these, twostage compressors are used, especially for larg
sizes, up to ratios of 34 to 51 (500 to 750 Ibs. gage). Some threestage con
pressors are used for ratios as low as 11 or 14 (150 or 200 Ibs. gage), althoug
installations of this nature are rare, and are warranted only when power is cost]
and the installation permanent and continuously used to warrant the hig
investment cost. As a minimum ratio for three stages, 11 (150 Ibs. gag*
is used for large units, while a few small units compress as high as 135 or eve
170 atmospheres (2000 or 2500 Ibs. gage). A notable use for the fourstaj
compressor is for charging the air flask of automobile torpedoes used by tl
various navies, which use pressures from 1600 to 3000 Ibs. per square inci
(110 to 200 atmospheres). These require special design of valves, cylindei
and packings to withstand the extremely high pressures, small clearances, ai
special precautions against leakage, due to the great loss of volumetric efficien
and economy that would otherwise result.
20. Work at Partial Capacity in Compressors of Variable Capacity,
is seldom that a gas compressor is run continuously at its full capacity.
the duty of the compressor is to charge storage tanks, it may be made to
at its full capacity until the process is completed and then may be stopp
entirely, by hand. Even where the compressed gas is being used continuou
it is common practice to have a storage reservoir into which the compressor m
deliver. This enables the compressor to deliver a little faster or slower tr
the demand for a short period without a great fluctuation pressure in
reservoir. For many purposes hand regulation is not sufficient or is
expensive, hence the demand for automatic systems of capacity regulati
These systems may be classified in a general way in accordance with the met!
WORK OF COMPRESSORS 161
)f driving. Some methods of power application permit of speed variations while
>thers require constant speed. The former provides in itself a means of regulat
ng capacity within certain limits, while, if the compressor must run at constant
ipeed, some additional means of gas capacity control must be provided.
Compressors driven by an independent steam engine, or steam cylinders
instituting part of the same machine may be made to run at any speed required
jvdthin a very wide range and still kept low enough for safety. If driven
gear, belt, rope, chain or direct drive from a source of power whose speed
s constant, the speed of the compressor cannot be varied. Electric motor,
5asengine, oilengine or waterpower drives are subject to only limited speed
ilteration and may, therefore, be placed in the constant speed class.
Regulation of Capacity by Means of Speed Change. If the speed of the com
pressor is decreased below normal:
1. Displacement of piston is decreased in proportion to the speed.
2. Mean effective pressure, as to hypothetical considerations, is the same,
due to the decrease of velocities in gas passages, the frictional fall of pres
jure during inlet and delivery is not so great, and hence the mean effective pres
ire is not quite so great. If the compressor is multistage, since a smaller
quantity of gas is passing through the intercooler, it is probable that the inter
jooling is more nearly perfect, thus decreasing the mean effective pressures in
?jbhe succeeding stages.
3. The volumetric efficiency is changed, due first to the fact that leakage
s about the same in total amount per minute as at full speed, but the total
juantity of gas being less, leakage is a larger percentage of the total; second,
,he inertia of gases in the supply pipe, as well as their friction, has been decreased.
The former tends to decrease vulmetric efficiency, while the latter may tend
,o increase or decrease it. It may be expected that the true volumetric efficiency
be somewhat greater at fractional speed than at full speed.
For any compressor there is a speed of maximum economy above and below
vhich the economy is less, though it may be that this most economical speed is
greater than any speed of actual operation.
It is not desirable at this point to discuss the effect of speed variation upon
(he economy of the engine or other motor supplying the power. The reasoning
foove applies to the term economy as applied to the compression effect obtained
jer unit of power applied in the compression cylinder. It might be noted here,
lowever, that the decrease of speed has little effect upon the mechanical efficiency
p the compressor as a machine, since frictional resistance between solid parts
iptemains nearly constant, and, therefore, power expended in friction will vary
(s the speed, as does approximately also the power to drive the compressor.
The ratio of frictional power to total may then be expected to remain nearly
Constant.
Regulation of Capacity at Constant Speed may be accomplished in a number
of ways :
1. Intermittent running;
2. Throttling the supply to compressor;
162 ENGINEERING THERMODYNAMICS
3. Periodically holding open or shut the intake valve;
4. Closing intake valve before end of intake stroke, or holding intake valve
open until compression stroke has been partially completed;
5. Large clearance;
6. Variable clearance,
The first necessitates some means for stopping and starting the compressor,
which is simple with electric drive, and may be accomplished in other cases by
means of a detaching clutch or other mechanical device. The pressure in the
reservoir is made to control this stopping and starting device by means of a
regulator. This arrangement is made to keep the pressure in the reservoir
between certain fixed limits, but does not maintain a constant pressure. The
economy of compression in this case is evidently the same as at full speed
continuous running, provided there is no loss in the driving system due to
starting and stopping, which may not be the case. This method of
regulation is used mainly for small compressors in which inertia is not
great, such as supply the air brakes on trolley cars. The sudden change
of load on the driving machinery would be too great if large compressors
were arranged in this way.
If the compressor whose capacity is regulated by intermittent running is
multitistage, the constant supply of water to the intercoolers while the compres
sor is stopped will lower the temperature of the cooling surface, causing more
nearly perfect intercooling when the compressor is started. Leakage, on the
other hand, will permit the loss of pressure to a greater or less extent in the
receivers while the compressor is stationary, which must be replaced after
starting before effective delivery is obtained.
Throttling the gas supply to the compressor has certain effects that may
be studied by referring to Fig. 39, which represents the hypothetical cycles most
nearly approaching this case. In order to reduce from the fullload lowpres
sure capacity, AB, to a smaller capacity, AE, the supply pressure is decreased
by throttling to the pressure of B', such that B' and E lie on the same adiabatic
The work area A'B'EA is entirely used up in overcoming the throttle resistance
and is useless friction, so that economy is seriously reduced by this methoc
of regulation. Such compressors may use almost as much power at partia
as at full capacity.
It is easily seen that this method of regulation would be undesirable, it
only advantage being simplicity.
The effect of throttling upon a multistage compressor may be illustrated a;
in Fig. 40, by considering the twostage compressor cycle without clearance
ABCDEF. The ratio of compression of the first cylinder is determined wit!
perfect intercooling by the ratio of displacements P c = P b (~\ . When th>
supply pressure is throttled down to P 6 ', the new receiver pressure will b
<Di\
jj } , a pressure much lower than P c . Hence the receiver pressur
is decreased, less work done in the first stage, and far more than half the wor
WORK OF COMPRESSORS
163
compression done in the second stage. If bestreceiver pressure existed at
normal capacity, it does not exist in the throttled condition.
The intake valve may be held wide open or completely closed during one
or more revolutions, thereby avoiding the delivery of any gas during that period.
If the intake valve is held wide open, the indicator card would be as shown in
Fig. 41A in full lines, ABCD, the dotted lines showing the cycle performed when
c'
i
i
A'
L.P.Cap Throttled >
L.P.Cap at Full Load
IG. 39. Effect of Throttling the Suction of Onestage Compressors, on Capacity and
Economy.
lormal operation is permitted. With the inlet valve open in this way there is
ib loss of power due to friction of the gas in passage during both strokes, measured
by the area within the loop.
Closing the inlet valve and holding it shut will give an indicator card of the
form EFG, Fig. 41 B, which will be a single line retraced in both directions
except for probable leakage effects. If leakage is small, there will be but little
164
ENGINEERING THERMODYNAMICS
area enclosed between the lines. At a high speed this might be expected
incur less lost power than the former plan.
Certain types of compressors are made with an intake valve controlled
by a drop cutoff, much like the steam valve of the Corliss engine. The effect
of this is to cut off the supply of gas before the end of the stroke, after which
time the gas must expand hypothetically according to the adiabatic law. The
return stroke causes it to compress along the same line continued up to the
delivery pressure, as indicated by the line PEG, Fig. 4 1C. There is little work
J
1
FIG. 40. Effect of Throttling Multistage Compressors on Receiverpressure, Work Distri
bution, Capacity and Economy,
lost in the process, none, if the line is superimposed as in the figure, and hence
the process is the same as if only the cycle AEGD were performed.
The same quantity of gas might have been entrapped in the cylinder by
holding the intake valve open until the end of the stroke and on the return till
the point E, Fig. 41D, was reached, then closing it. The same compression
line EG will be produced. The line AB will not coincide with BE, due to
friction of the gas in passages, and hence will enclose between them a small
area representing lost work, which may be no larger than that lost in the process
EFE.
WORK OF COMPRESSORS
165
If such an automatic cutoff were applied only to the first stage of a multi
stage compressor, the effect would be to lower receiver pressures as in the throt
tling process. To avoid this, the best practice is to have a similar cutoff to act
on the supply to all of the stages. If this is properly adjusted, the receiver
pressures can be maintained the same as at full load. An additional advantage
of this system is that even if the compressor is to be used for a delivery pres
sure for which it was not originally designed, the relative cutoffs may be so
adjusted as to give and maintain bestreceiver pressure.
1 *
l
\
\
\
1
I
\
\
v
\
\
\
\
\
N >
x
\
\
D
" > *^,
,^_
(Su
x Pr.
A
V
B
C
B
(A)
D G
\
\
(B)
C
D
FIG. 41. Control of Compressor Capacity by A. Open Inlet Valve; B. Closed Inlet Valve ;
C. Suction Cutoff; D. Delayed Suction Closure.
Since the lowpressure capacity per cycle of a compressor involves clearance
and ratio of compression as two of its variables, it is possible to change capacity
by changing either the clearance or the ratio of compression. fc
(L. P.
(227)
Assuming that clearance is a fixed amount and not zero, it is evident that an
increase in the ratio of compression decreases the capacity, and when it has
166
ENGINEERING THERMODYNAMICS
reached a certain quantity will make the capacity zero.
i
If the clearance is
large, making the coefficient of R p s large in the equation the effect of a change
in that factor is increased. Fig. 42 indicates the hypothetical performance
of a compressor with large clearance. When the pressure of delivery is low
(say P e ) the capacity is large, AB. The cycle is then A BCD. An increase
of the delivery pressure to P c f changes the cycle to A'BC'D' and the low pres
sure capacity is A'B. If the compressor is delivering to a receiver from which
no gas is being drawn, the delivery pressure will continue to rise and the capacity
FIG. 42. Variation of Compressor Capacity with Rise of Delivery Pressure, Fixed Clearance,
Pressure for Zero Delivery.
to decrease till the capacity approaches zero as the delivery pressure approaches
the pressure P e as a limit.
(limiting del.pr.) =(sup.pr.)(~ j ..... (228)
:. \ C I
When the limiting condition has been reached and the capacity has become
zero, the compression and reexpansion lines coincide and enclose zero area
between them; hence, the mean effective pressure and the indicated horse
power are zero," for the hypothetical case. Leakage will prevent a perfect
coincidence of the lines and cause some power to be required in addition to that
of friction.
Such a simple method of regulation as this is used for some small com
pressors driven constantly from some source of power used primarily for
other purposes. When it is not necessary to have a constant delivery pressure,
WORK OF COMPRESSORS
167
J but only to keep it between certain limits, this may be made use of, especially
if the limits of pressure are quite wide.
The expression for lowpressure capacity Eq. (227), suggests the possi
bility of decreasing capacity by the increase of clearance. The effect of this is
shown in Fig. 43. The original compression cycle (full capacity) is shown by
ABCD, with a clearance volume of cD, so that the axis of zero volume is OP.
Increasing the clearance to c'D causes a smaller volume C'D to be delivered
and due to the more sloping reexpansion DA', a smaller volume of gas is
taken in, A 'B.
It has been shown in previous sections that clearance has no effect upon the
economy of a compressor so far as hypothetical considerations are regarded.
In practice it is found that a slight loss of economy is suffered at light load,
o'
P D
C' C
\
L.P. Cap Full Load
D
L.P. Cap Part Load
FIG. 43. Variation of Compressor Capacity by Changing Clearance.
as might be expected, due to greater leakage per unit of capacity. The addi
tional clearance is provided in the form of two or more chambers connected to the
clearance space of the compressor by a passage in which is a valve automatically
controlled by the receiver pressure.
In the multistage compressor, decreasing the capacity of the first stage by
an increase of its clearance would evidently permit a decrease of receiver pres
sures unless the capacity of each of the various stages is decreased in the same
proportion. Eq. (132) gives the condition which must be fulfilled to give best
receiver pressure for a twostage compressor.
J_1
2*
168
ENGINEERING THEEMODYNAMICS
Since Di, Z>2, and R p remain fixed, for any chosen value of clearance of the
first stage, ci, the clearance of the second stage, C2, to give bestreceiver
pressure can be found,
C 2 =
(Rp2~sl)R p 2sD 2
(229)
For every value of firststage clearance there is a corresponding clearance of
second ^stage that will give bestreceiver pressure, found by this equation. Sim
ilar reasoning can be applied to three or fourstage compressors.
2i. Graphic Solution of Compressor Problems. In order to obviate the
necessity of working out the formulas given in this chapter each time a prob
lem is to be solved, several of them have been worked out for one or more
cases and results arranged to give a series of answers graphically. By the
use of the charts made up of these curves many problems may be solved
directly and in many others certain steps may be shortened. A description of
each chart, its derivation and use is given in subsequent paragraphs.
Chart, Fig. 44. This chart gives the work required to compress and deliver
a cubic foot of (sup.pr.) air or the horsepower to compress and deliver 1000 cu.ft.
of (sup.pr.) air per minute if the ratio of pressure (del.pr.)^ (sup.pr.), the value
of s and the (sup.pr.) are known, and compression occurs in one stage. The
work or H.P. for any number of cubic feet is directly proportional to number
of feet. The curves are dependent upon the formulas, Eq. (31), for the case
when 8 = 1, and Eq. (51) for the case when s is not equal to 1. These formulas
are:
Eq. (31), W per cu.ft. = 144 (sup.pr) log e R p ',
(51), W per cu.ft. = 144  (sup.pr.)
sl
These equations are difficult to solve if an attempt is made to get a relation
between the work and ratio of pressures. This relation may, however, be
worked out for a number of values of pressure ratios and results plotted to
form a curve by which the relation may be had for any other ratio within
limits. This has been done in this figure in the following manner:
On a horizontal base various values of R p are laid off, starting with the value
2 at the origin. The values for work were then found for a number of values
of R p with a constant value of (sup.pr.) and s. A vertical work scale was
then laid off from origin of R p and a curve drawn through the points found
by the intersection of horizontal lines through values of work, with vertical
lines through corresponding values of R p . The process was then repeated for
other values of s and curves similar to the first, drawn for the other values
of s. From the construction so far completed it is possible to find the work per
cubic foot for any pressure ratio and any value of s for one (sup.pr.) by pro
WORK OF COMPRESSORS
169
UT bs aad 'sqi ('JcT 'dns)
17 o ENGINEERING THERMODYNAMICS
WORK OF COMPRESSORS
171
CO *. 1
o
I
I!
o
I
o
I
172
ENGINEERING THEKMODYNAMICS
I . . .
SJ 3$ 8
sqy *ui nbs aad ajy ('Jd 'dn
WORK OF COMPRESSORS 173
C er to (sup.pr.) = 15 and down, and the horsepower will be found to be 13.6
4 before by use of formulas.
Chart, Fig. 47. This chart is for finding the (m.e.p.) of compressors. In the
c?e of multistage compressors with bestreceiver pressure and perfect inter
pling, the (m.e.p.) of each cylinder may be found by considering each cylinder
4 a singlestage compressor, or the (m.e.p.) of the compressor referred to the
P. cylinder may be found.
The chart depends on the fact that the work per cubic foot of (sup.pr.) gas
equal to the (m.e.p.) for the no clearance case and that the (m.e.p.) with
arance is equal to the (m.e.p.) for no clearance, times the volumetric
iciency. Diagrams 1, 2 and 4 are reproductions of Figs. 45, 46 and 47 to a
aller scale and hence need no explanation as to derivation. Their use may
briefly given. From the proper ratio of pressures project upward to the
Dper curve, then horizontally to the (sup.pr.) and downward to read work per
bic feet of (sup.pr.) gas.
The volumetric efficiency diagram was drawn in the following manner :
om Eq. (65) vol. eff. = (l+c cR p s ), showing that it depends upon three
riables, R P , c and s. A horizontal scale of values of. R p was laid off. Values
i
R p s were found and a vertical scale of this quantity laid off from the same
gin as the R P values. Through the intersection of the verticals from various
lues of R p with the horizontals drawn through the corresponding values of
P) S for a known value of s, a curve of this value of s was drawn. In a similar
y curves of other values of s were drawn. From the construction so far
i
aipleted it is possible to find the value of (R p )
174
ENGINEERING THERMODYNAMICS
f 1 8 9 10 11 12 13 14
tfatio of Pressures
81 77 70 63 5fe 49 4\i 35 28 21 14 7
Work per Cu. Ft. of (Sup. Pr.) GasH44
FIG.
f f
'7 70 63
Work
47. Mean Effective Pressure of Compressors, One, Two, and Threestages.
WORK OF COMPRESSORS
Initial Pressure Lbs. per Sq. In. Abs.
175
11 12
14 15
7 8 9 10
Ratio of Pressures
70 63 56 49 42 35 28 21 14 7
AVork per Cu. Ft. of (Sup. Pr.) GasM44
FIG. 47. Mean Effective Pressure of Compressors, One, Two, and Threestages.
176 ENGINEERING THERMODYNAMICS
each to be a singlestage compressor and remembering that (1 rec.pr.) becomes
(sup.pr.) for second stage, and (del.pr.) for first stage and that (2 rec.pr.)
becomes (sup.pr.) for third stage, (del.pr.) for second stage. The (m.e.p.) reduced
to lowpressure cylinder is found by taking work per cubic feet of (sup.pr.)
gas and multiplying by volumetric efficiency of lowpressure cylinder.
To illustrate the use of this curve the example of Section (16) may be
solved. Projecting upward from the pressure ratio of 9.35 to the line of s = lA
and then over to (sup.pr.) = 15 in diagram 4, since compression is three stage
and from 15 Ibs. per square inch to 140 Ibs. per square inch, work per cubic
foot or (m.e.p.), is found for no clearance to be 37.8 abs. per square inch. Since
bestreceiver pressure assumed is 31.6, which gives a ratio of 2.1 for the low
pressure cylinder. From diagram 3, by projecting upward from R P = 2.I and
over to the 5 per cent clearance line volumetric efficiency is 96.5. The product
(m.e.p.)Lan
gives (m.e.p.)reduced to lowpressure cylinder and is 36.5. From the
formula, horsepower is found to be 358 as before.
Chart, Fig. 48. As mentioned in Section 18, there is one (sup.pr.), which
for a given (del.pr.) will give the maximum work of compression. The chart,
Fig. 48, originated by Mr. T. M. Gunn, gives a graphical means of finding this
value of (sup.pr.) when the (del.pr.), clearance and value of s are known. It also
gives on the righthand of the chart a means for finding the (m.e.p.) for this
condition. The figure was drawn by means of Eqs. (214) and (218). For the
value of s = l the ratio of (del.pr.) to (sup.pr.) was found for cases of clearances
from per cent to 15 by means of Eq. (218) by trying values of this ratio which
/del.pr. \ c /del.pr. \
would fulfill the condition of the equation, logel p ) = ~jqL~( gup pr ).
For values of s not 1, Eq. (214) was used, and a set of values of R p found
for the values of s = 1.4 and 1.2 by trial, the correct value of R p being that which
satisfied the equation,
/deLpr
Vsup.
pr\ = __ [" __ s1 /del.pr. \ .1
.pr./ 1+c L ~ * \suPpr. / J
As an example the work for the case where s = 1.2 and c = 10 per cent is given.
Try fl, = 2.6, then, R^ =
= 1.091(1. 1.01667X2.218),
= 1.161
WORK OF COMPRESSORS
\
8 # 8 $
(d*a*tn) jo OT:VBH mmnproH*
'saanssaajjo ORBH
178
ENGINEERING THERMODYNAMICS
R p = (1.161) 6 = 2.45, which shows the value of 2.6 to be incorrect. For a second
trial take 2.45, and then,
,*, = 1.091(1.1  .01667 X2.45' 833 ),
= 1.1627
= (1.1617)6 = 2.458,
which is sufficiently close. Therefore the value of R p for s = 1.2 and clearance =
10 per cent, is 2.45; that is, maximum work will occur for a given (del.pr.)
when (sup.pr.) is ^5 times the ( del P r )
When the values for R P had been obtained a horizontal axis of values of s
and a vertical one of R P values, were laid off and the points for clearance curves
laid off to their proper values referred to these axes. Through points as plotted
the clearance lines were drawn. The righthand diagram was plotted in a
similar manner from Eqs. (213) and (217), for s not equal to 1 and equal to
1 respectively.
The latter formula was rearranged in the form
(del.pr. / " R p * E "
the last term being found from curve of Fig. 45. The value of R p for each
value of the clearance was taken from the lefthand diagram, and substituted
in the above expression to obtain ( rp L Jfor the case of s = 1. Eq. (213) was
put in form
(m.e.p. \ _ *
deLpr./ (*
and values of R p for each value of trie clearance found in the lefthand diagram
were substituted, together with E v values from Fig. 45 and the value of ( , ', ' j
found for each case of clearance when s = 1.4. When the points for s = l and
s = 1.4 had been found, a horizontal axis of values of s and a vertical one of
values of R p were laid off, and points for the clearance curves plotted as for
the lefthand diagram and the curves drawn in.
To find the (sup.pr.) to give maximum work for any (del.pr.) it is only
necessary to project from the proper value of s to the proper clearance curve,
and then horizontally to read the value of R p . The (del.pr.) divided by this
gives the (sup.pr.) desired. To obtain the (m.e.p.) project upward from the
value of s to the clearance curve, then horizontally to read the ratio ( rr I
\ del.pr./
The (del.pr.) times this quantity gives the m.e.p.
As an example of the use of this chart let it be required to find the (sup.pr.)
for the case of maximum work for 9X12 in. doubleacting compressor running
200 R.P.M., having 5 per cent clearance and delivering against 45 Ibs. per square
inch gage. Also the horsepower. Compression such that s = 1.3.
WORK OF COMPRESSORS
179
Projecting from the value 1.3 for s on the lefthand diagram to the line of
5 per cent clearance find R p to be 2.8, hence (sup.pr.) = 2^ = 21.4 Ibs. per square
inch absolute = 6.4 Ibs. per square inch gage. Again, projecting from value 1.3
for s on righthand diagram to line of 5 per cent clearance find that;', 6 '*; = .383,
(del.pr.)
hence (m.e.p). = 23 and
I.H.P.
23X1X64X400
= 17.8.
1.00
IM
II
I
ft .80
I
o .78
.71
.72
S=l.l
S =
8 = 1.3
8 = 1.4
8 = 1.3
8 = 1.5
8=1.4
8 = 1.5
2 3 4 5 67 8 9 10 11 12 13 14 15
Ratio of Pressure = Rp
Note; Solid Lines = 3 Stage; Broken Lines = 2 Stage
FIG. 49. Relative Work of Two and Threestage Compressors Compared to Single Stage.
180 ENGINEERING THERMODYNAMICS
The chart was made by laying off on a horizontal base a scale of pressure
ratios. From the same origin a scale of work for two or three stage divided by
the work of one stage was drawn vertically. For a number of values of R P
the work to compress a cubic foot of gas was found for one, two and three stage
for each value of s. The values found by dividing the work of two or three
stage by the work of single stage were plotted above the proper R p values and
opposite the proper ratio values and curves drawn through all points for one
value of s. To find the saving by compressing in two or three stages project
from the proper R p value to the chosen s curve for the,desired number of stages,
then horizontally to read the ratio of multistage to onestage work. This value
gives per cent power needed for one stage that will be required to compress
the same gas multistage. Saving by multistage as a percentage of single
stage is one minus the value read.
To illustrate the use of this chart, find the per cent of work needed to
compress a cubic foot of air adiabatically from 1 to 8J atmospheres in two
stages Compared to doing it in one stage. From examples under chart Nos.
44 and 46 it was found that work was 6300 ft.lbs. and 5320 ft.lbs. respec
tively, for one and twostage compression, or that two stage was 84.5 per cent
of one stage. From R p , 8J project up on Fig. 49 to s = 1.406 for two stage and
over to read 84.6 per cent, which is nearly the same.
Chart, Fig. 50. This chart, designed by Mr. T. M. Gunn, shows the
economy compared to isothermal compression.
The chart was drawn on the basis of the following equation:
Economy (isothermal) = g^.J!thermaJ (no clearance)
m.e.p. actual rE v actual
(sup.pr.)
sl
^r (sup.pr.) (R p s 1)
o J.
\OgeRp
Values of this expression were worked out for each exponent, for assumed values
of R p . A scale of values of R p was laid off horizontally and from the same
origin a vertical scale of values of the ratio of isothermal to adiabatic. The
results found were then plotted, each point above its proper R p and opposite its
ratio value. Curves were then drawn through all the points found for the
same value of s. In a similar way a set of curves for two stage and a set for
three stage were drawn.
This chart is also useful in obtaining the (m.e.p.) of the cycle if the (sup.pr.)
and the volumetric efficiency of the cylinder be known. A second horizontal
scale laid off above the R p scale shows the (m.e.p.) per pound of (sup.pr. for)
the isothermal noclearance cycle. This is found to be equal to log e R P} since
WORK OF COMPRESSORS
r .
qgiiAv pamluioo ^uiouoo^
182
ENGINEERING THERMODYNAMICS
the (m.e.p.) for no clearance is equal to the work per cubic foot of (sup.pr.) gas,
which, in turn, for the isothermal case is (sup.pr.) log* R p or loge R v when
(sup.pr.) = 1.
Knowing the ratio of pressures, economy compared to isothermal can be
found as explained above. Also knowing R p the (m.e.p.) per pound initial is
found from the upper scale.
Since the latter quantity is assumed to be known, by multiplying it by
factor just found there is obtained (m.e.p.) isothermal. Since volumetric efficiency
is assumed known, all the factors are known for the first equation given above
which, rearranged, reads
. (m.e.p.) actual =
m.e.p. isothermal (no clearance)
(economy isothermal) f E v
Chart, Fig. 51. This chart is drawn to give the cylinder displacement for a
desired capacity, with various values of R p , s and clearance. From the formula
Eq. (64)
(L. P. Cap.)=Z>(l+cc# p *).
The righthand portion of the diagram is for the purpose of finding values
of (Rp) * for various values of R p and s, and is constructed as was the similar
curve in Fig. 45. The values of the lower scale on the lefthand diagram give
JL
values of D = (L. P. Cap.)J(l+c cR p s ), where capacity is taken at 100 cu.ft.,
this scale was laid out and the clearance curves points found by solving the
j_
above equation for various values of (R p ) s for each value of c. To obtain the
displacement necessary for a certain capacity with a given value of R PJ c and
s, project upward from R p to the proper s curve across to the c curve and down to
read displacement per hundred cubic feet. Also on the lefthand diagram are
drawn lines of piston speed, and on lefthand edge a scale of cylinder areas
and diameters to give displacements found on horizontal scale. To obtain
cylinder areas or approximate diameters in inches project from displacement to
piston speed line and across to read cylinder area or diameter. Figures given
are for 100 cu.ft. per minute. For any other volume the displacement and
area of cylinder will be as desired volume to 100 and diameters will be as
Vdesired volume to 100.
As an example, let it be required to find the lowpressure cylinder size for a
compressor to handle 1500 cu.ft. of free air per minute. Receiver pressure to
be 45 Ibs. per square inch gage and (sup.pr.) to be atmosphere. Piston speed
limited to 500 ft. per minute. Compression to be so that s = 1.4 and clear
ance =4 per cent. Projecting upward from ft p = 4 tos = 1.4, across to c = 4%,
and down to piston speed = 500, find the diameter of a cylinder for 100 cu.ft.
per minute is 6.3. For 1500 cu.ft. diameter will be as Vl5X6.3 = 3.9X6.3 =
ins.
WORK OF COMPRESSORS
183
8 4 ENGINEERING THERMODYNAMICS
GENERAL PROBLEMS ON CHAPTER II.
Prob. 1. One hundred cubic feet of H 2 S are compressed from 15 Ibs. per square
ich absolute to 160 Ibs. per square inch absolute.
(a) Find work done if compression occurs isothermally in a noclearance onestage
compressor;
(6) Adiabatically in a twostage, noclearance compressor;
(c) Adiabatically in twostage compressor each cylinder having 5 per cent clearance ;
Prob. 2. Air is being compressed in three plants. One is singlestage, the second is
srostage, and the third is threestage Considering the compressors to have no clear
ace and 1000 cu.ft. of free air compressed adiabatically per minute from atmosphere
3 150 Ibs. per square inch gage, what will be the horsepower required and cylinder
Lzes in each case?
Prob. 3 A twostage compressor with 5 per cent clearance n the h'gh and 3 per
ent in the lowpressure cylinder is compressing air from 14 Ibs. per square inch ab
olute to 125 Ibs. per square inch gage. What is the bestreceiver pressure and what
lust be the size of the cylinders to handle 500 cu.ft. of free air per minute?
Prob. 4. A manufacturer gives a capacity of 970 cu.ft. for a 22 J X24 in. single
tage compressor running at 142 R.P.M. when working pressures are 50 to 100 Ibs. per
quare inch gage. What would be the clearance for each of these pressures assuming
= 1.4?
Prob. 5. The card taken from a singlestage compressor cylinder showed an appar
nt volmetric efficiency of a 95 per cent and a mean value for s of 1.3. What is the
learance and what would be the (m.e.p.) for the ratio of pressures of 6?
Prob. 6. A compressor with doubleacting cylinder 12x14 ins., having 6 per cent
learance, is forcing air into a tank. Taking the volumetric efficiency as the mean of
hat at the start and the end, how long w 11 it take to build up 100 Ibs. per square inch
;age pressure in a tank of 1000 cu.ft. capacity if the speed is 100 M.R.P. and compres
ion is isothermal, and how long wi 1 it take to do it if the compression is adiabatic and
he ah* in the tank does not cool during filling? What is the maximum attainable pressure?
Prob. 7. It is desired to supply 1000 cu.ft. of air per minute at a pressure of 80 Ibs.
>er square inch gage. A twostage compressor is to be used, the clearance in low pres
ure of which is 3 per cent. What must be the displacement of the lowpressure cylin
ler and what will be the horsepower of the compressor?
Prob. 8. The lowpressure cylinder of a compressor is 18x24 ins. and has a clear
tnce of 4 per cent. The receiver pressure is 60 Ibs. per square inch absolute. The high
>ressure cylinder has a 5 per cent clearance. What must be its diameter, stroke being
ame as low, so that compressor will operate at its designed receiver pressure?
Prob. 9. The discharge pressure of a twostage compressor is 120 Ibs. per square
och absolute and the supply pressure is 15 Ibs. per square inch absolute The compressor
s 10x16x12 ins. The clearance in the lowpressure cylinder is 3 per cent. What
Qust be the clearance in the highpressure cylinder for the machine to operate at best
eceiver pressure? Compression is adiabatic.
Prob. 10. If the clearance in the highpressure cylinder df Prob. 9 were reduced
o 3 per cent, would the receiver pressure increase or decrease, how much and why?
Prob. 11. If the discharge pressure in Prob. 9 fell to 100 Ibs. per square inch absolute,
vhat would be the new bestreceiver pressure and why? Would the original clearance
illow the new bestreceiver pressure to be maintained?
PROBLEMS ON CHAPTER II 185
Prob. 12. The discharge pressure for which a 20 Jx 32^x24 in. compressor is
lesigned, is 100 Ibs. per square inch gage, supply pressure being 14 Ibs. per square inch
Absolute. The d'scharge pressure is raised to 125 Ibs. per square inch gage. The
ilearance on the highpressure cylinder can be adjusted. To what value must it be
;hanged to enable the compressor to carry the bestreceiver pressure for the new
lischarge pressure? Lowpressure clearance is 5 per cent at all times and corn
session being adiabatic.
Prob. 13. A manufacturer builds his 15^X25^X18 in. compressors with lowpres
ure cylinders of larger diameter for high altitude work. What would be the diameter
>f a special cylinder for this compressor to work at an altitude of 10,000 ft. and what
v r ould be the horsepower per cubic foot of lowpressure air in each case?
Prob. 14. A threestage compressor has 4 per cent clearance in all the cylinders. The
owpressure cylinder is 34x36 ins., delivery pressure 200 Ibs. per square inch gage,
upply pressure 14 Ibs. per square inch absolute. What must be the size of the other
Cylinders for the machine to operate at bestreceiver pressure.
Prob. 15. The cylinders of a twostage compressor are given as 10J and 16^ ins.,
he stroke being 12 ins. The machine has a capacity of 440 cu.ft. per minute at 160
I.P.M., the supply pressure is 14 Ibs. per square inch absolute and the delivery
pressure 100 Ibs. per square inch gage. What is the clearance of each cylinder?
Prob. 16. To perform the required useful refrigerating effect in an ice plant, it is
iccessary to compress 250 cu.ft. of ammonia vapor per minute from 30 Ibs. per square
ach gage to 150 Ibs. per square inch gage. What must be the size of the compressor
o handle this at 75 R.P.M. if it be double acting and the clearance be 5 per cent?
Prob. 17. The maximum delivery pressure of a type of compressor is controlled by
aaking the clearance large so that the volumetric efficiency will decrease as the pressure
ises and become zero at the desired pressure. What must be the clearance for a single
tage compressor where the supply pressure is 14 Ibs. per square inch absolute and the
aaximum delivery pressure 140 Ibs. per square inch absolute? What will be the volu
aetric efficiency of the same machine at a delivery pressure of J the maximum? At ?
Prob. 18. A threestage compressor has a clearance of 5 per cent in each cylinder.
Vhat must be the cylinder ratios for the bestreceiver pressures when the machine is
ompressing to 170 Ibs. per square inch gage from atmosphere?
Prob. 19. Show why it was very essential to keep the clearance low in cylinders of
hreestage compressor used for compressing air for airdriven cars, where the delivery
ressure carried was 2500 Ibs. per square inch, by assuming numerical data and
alculating numerical proof.
Prob. 20. With water falling 150 ft. and used to compress air directly, how many
ubic feet of air could be compressed per cubic foot of water?
Prob. 21. Air is compressed from atmosphere to 150 Ibs. per square inch absolute,
lothermally in one stage. How much more work would be required per cubic foot if
ompression were adiabatic? How much of this excess would be saved by compressing
wo stage? Three stage?
Prob. 22. 150 I.H.P. is delivered to tluTair cylinders of a 14 X22i Xl8 in. compres
)r, running at 120 R.P.M. The supply pressure is 15 Ibs. per square inch absolute,
'he volumetric efficiency as found from the indicator card is 95 per cent. What was
e discharge pressure?
Prob. 23. The clearance in the highpressure cylinder of a compressor is 5 per cent,
hich allows the compressor to run with the bestreceiver pressure for a discharge of
00 Ibs. per square inch absolute when the compressor is at sealevel. What would the
L86 ENGINEERING THERMODYNAMICS
clearance be if the discharge pressure were kept the same and the altitude were 10,000
ft. to keep the bestreceiver pressure?
Prob. 24. How many cubic feet of supplypressure air may be compressed per minute
from 1 to 8 atmospheres absolute by 100 horsepower if the compression in all cases is
adiabatic?
(6) Three stage, no clearance;
(c) Two stage with 5 per cent clearance;
(d) Single stage with 5 per cent clearance;
(a) Two stage, no clearance
Prob. 25. The capacity of a 14iX22ixl4 in. compressor when running at 140
R.P.M. is said to be 940 cu.ft. for working pressures of 80 to 100 Ibs. per square inch
gage and atmospheric supply at sea level. Check these figures.
Prob. 26. What horsepower would be required by an 18ix30ix24 in. compressor;
operating at 100 P.P.M. and on a working pressure of 100 Ibs. per square inch gage if;
the clearance in lowpressure cylinder is 4 per cent? What would be the capacity?
Prob. 27. By means of water jackets on a compressor cylinder the va ue for s of comj
pression curve in singlestage machine is lowered to 1.3. Compare the work to com]
press 1000 cu.ft. of air from 1 to 8 atmospheres absolute with this condition with thaj
required for isothermal and adiabatic compression.
Prob. 28. What must be the size of cylinders in a threestage compressor for coml
pressing gas from 50 Ibs. per square inch absolute to 600 Ibs. per square inch absolute
when s equals 1.3 and each cylinder has 3 per cent clearance. Compressor to run ai
100 R.P.M. to be double acting and handle 1000 cu.ft. of gas per hour?
Prob. 29. How many cu.ft. of free air could be compressed per minute with an avail
able horsepower of 1000 H.P. from atmosphere to 150 Ibs. per square inch gage; (a)
if compression is isothermal; (6) if compression be singlestage adiabatic; (c) if com
pression be threestage adiabatic?
Prob. 30. A singlestage compressor is compressing air adiabatically at an altitude
of 6000 ft. to a pressure of 80 Ibs. per square inch gage. The cylinder has 2 per cent
clearance. What must be the s'.ze of the cylinder to compress 2000 cu.ft. of free air pel
minute if the piston speed is limited to 600 ft. per minute? What, if the clearance be
zero?
Prob. 31. What would be the size of the twostage compressor for same data as in
Prob. 30?
CHAPTER III
VORK OF PISTON ENGINES. HORSE POWER AND CONSUMPTION OF
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER
GAS OR VAPOR UNDER PRESSURE.
1. Action of Fluid in Single Cylinders. General Description of Structure
ind Processes, The most commonly used class of engines is that in which
,he operation is dependent on the pushing action of highpressure fluids on
)istons in cylinders, and this includes all piston steam engines of the recip
ocating or straightline piston path group as well as the less common
otary group, having pistons moving in curved and generally circular paths.
Cn these same engines there may be used compressed air as . well as steam,
md equally as well the vapors of other substances or any other gases, without
shange of structure, except perhaps as to proportions, providing only that the
substance to be used be drawn from a source of supply under high pressure,
admitted to the cylinder, there used and from it discharged or exhausted
;o a place of lower pressure. This place of lower pressure may be the open
iir or a closed chamber; the used fluid may be thrown away and wasted or
ised again for various purposes without in any way affecting the essential
Drocess of obtaining work at the expense of highpressure gases or vapors.
[t is evident that, regarding a piston as a movable wall of a cylinder, when
3ver a fluid acts on one side with greater pressure than on the other, the
Diston will move toward the lower pressure end of the cylinder, and in so
noving can exert a definite force or overcome a definite resistance, measured
the difference in pressure on the two sides and the areas exposed to the
Dressure. It is not so evident, but just as true, that the piston may be made
io move from one end of the cylinder to the other when the average pressure
one side is greater than the average pressure on the other, and also do work
3ven if the excess of pressure should reverse in direction during the stroke
provided only some energy storage device is added. In the common
steam or compressedair engine this is a flywheel with the usual connecting
"od and crank mechanism, uniting the reciprocating piston movement with
ihe continuous rotary movement of the flywheel mass. In certain forms
pumps the energy is stored in extra cylinders at times of excess and given
Nit at times of deficiency in the path of the piston, so that its motion from
to end of cylinder may not be interrupted even if the pressure on the
driving side should fall below that on the resisting side, assuming, of course,
he average pressure for the whole stroke to be greater on the driving side
ban on the resisting side.
187
188
ENGINEERING THERMODYNAMICS
It appears, therefore, that piston movement in engines of the common
form and structure, and the doing of work by that movement is not a
question of maintaining a continuously greater pressure on one side than
on the other. On the contrary, the process is to be studied by examination
of the average pressure on the driving side and that on the resisting side,
or by comparing the whole work done on one side with the whole work
done on the other side by the fluid. The work done by the fluid on one
side of a piston may be positive or negative, positive when the pressures are
assisting motion, negative when they are resisting it. It is most con
venient to study the action of fluids in cylinders by considering the
whole action on one side from the beginning of movement at one end to
the end of movement at the same point, after the completion of one complete
forward and one complete return stroke. All the work done by the pressure
of the fluid on the forward stroke on the side of the piston that is apparently
moving away from the fluid is positive work, all the work done by the pres
sure of the fluid on the same side of the piston during the return stroke is
negative, and for this stroke the side of the piston under consideration is
apparently moving toward the fluid or pushing it.
For the complete cycle of piston movement covering the two strok
the work done on one side is the algebraic sum of the forward strok'
work, considered positive, and the back stroke work, considered negative.
During the same time some pressures are acting on the other side of the
piston, and for them also there will be a net work done equal to the cor
responding algebraic sum. The work available for use during the comple
two strokes, or one revolution, will be the sum of the net work done by th
fluid on the two sides of the piston during that time, or the algebraic sum o
two positive and two negative quantities of work. Methods of analysis of
work of compressed fluids in cylinders are consequently based on the action i
one end of a cylinder, treated as if the other end did not exist.
Just how the highpressure fluid from a source of supply such as a boiler
or an air compressor is introduced into one end of a cylinder, how it is treated
after it gets there, and how expelled, will determine the nature of the varia
tion in pressure in that end that acts on that side of the piston, and th
are subjects to be studied. To determine the work done in the cylinde
end by the fluid, it is necessary to determine laws of pressure change with
stroke, and these are fixed first by valve action controlling the distribution
of the fluid with respect to the piston and second by the physical properties
of the fluid in question.
It is necessary that the cylinder be fitted with a valve for getting fluid
into a cylinder, isolating the charge from the source of supply and getting i
out again, and it may be that one valve will do, or that two or even more
are desirable but this is a structural matter, knowledge of which is assumed
here and not concerned with the effects under investigation. The first ste,
in the process is, of course, admission of fluid from the source of supply to th
cylinder at one end, which may continue for the whole, or be limited to a
WOEK OF PISTON ENGINES 189
3f the stroke. When admission ceases or supply is cut off before the end
 3f the stroke there will be in the cylinder an isolated mass of fluid which
will, of course, expand as the piston proceeds to the end. Thus the forward
190
ENGINEERING THERMODYNAMICS
may evaporate; the resistance through valves will always make the cylinder
pressure during admission less than in the supply chamber and greater during
exhaust than the atmosphere or than in exhaust receiver or condenser and may
through the valve movements make what might have been a constantpres
sure straight line become a curve. There will, by reason of these influences,
encountered in real engines, be an almost infinite variety of indicator cards
or pressurevolume cycles for such engines.
\
AB STEAM ADMITTED
BC " EXPANDED
CD EXHAUSTED
DA " COMPRESSED
A'
FIG. 52. Diagram to Indicate Position of Admission, Cutoff, Release, Compression on)
Engine Indicator Card.
The various points of the stroke at which important events occur,,
important in their pressurevolume significance, have names, as do also the
lines between the points, and these names are more or less commonly accept
and generally understood as follows: letters referring to the diagram Fig. 52.
Point Names: Events of Cycle.
A. Admission is that point of the stroke where the supply valve i$
opened.
B. Cutoff is that point of the stroke where the supply valve is closed.
C. Release is that point of the stroke where the exhaust valve is opened.
D. Compression is that point of the stroke where the exhaust vab
is closed.
Names of Lines, or Periods:
iB. Admission or steam line joins the points of admission and cutoff]
BC. Expansion line joins the points of cutoff and release.
CD. Exhaust line joins the points of release and compression if th<
is any, or admission if there is not.
DA. Compression line joins the points of compression and admission
WORK OF PISTON ENGINES
191
By reason of the interferences discussed, these points on actual indicator
ards may be difficult to locate, one line merging into the next in so gradual
manner as to make it impossible to tell where the characteristic point lies,
^s will be apparent from Fig. 53, in which is reproduced a number of actual
ndicator cards. In such cases equivalent points must be located for study
KG. 53. Actual Steam Engine Indicator Cards Showing Distortion of Lines and Uncertain
Location of Characteristic Points.
These same terms, which it appears sometimes refer to points and some
Imes to lines, are also used in other senses, for example, cutoff is com
Qonly used to mean the fraction of stroke completed up to the point of
utoff, and compression that fraction of stroke remaining incomplete at the
int of compression, while compression is also sometimes used to express
.he pressure attained at the end of the compression line. In general, there
192
ENGINEERING THERMODYNAMICS
is nothing in the use of these words to indicate just which of the various
meanings is intended except the text, and experience will soon eliminate most
of the possible chances of confusion.
Prob. 1. Draw a card in which admission and exhaust are late. Draw a card in
which there is no compression, in which compression is very early, so that compression
pressure is equal to admission pressure. Draw a card with per cent cutoff, and cut
off = 100 per cent. Draw cards with same cutoff but with varying initial pressures.
Prob. 2. The following diagrams, Fig. 54, are reproductions of indicator care
actually taken from engines. Explain what features are peculiar to each and
possible give an explanation of the cause.
No. 3
No. 4
No. 5
No. 7
No. 8
Fi<;. 54. Indicator Diagram from Steam Engines with Improperly Set Valve Gear.
For example, in No. 1 a line of pressure equalization between the end of the com
sion line and the beginning of the admission line inclines to the right instead ol
being perpendicular, as in a perfect diagram. This is due to the fact that admis
ion does not occur until after the piston has begun to move outward, so that pres
t occur at constant volume, but during a period of increasing volume
2. Standard Reference Cycles or PV Diagrams for the Work of Expansive
Fluids in a Single Cylinder. Simple Engines. To permit of the derivatioi
: a formula for the work of steam, compressed air, or any other fluid in
WORK OF PISTON ENGINES 193
jylinder, the various pressure volume, changes must be defined algebraically.
The first step is, therefore, the determination of the cycle or pressurevolume
liagram representative of the whole series of processes and consisting of a
lumber of wellknown phases or single processes. These phases, ignoring all
sorts of interferences due to leakage or improper valve action, will consist of
constantpressure and constantvolume lines representing fluid movement
into or from the cylinder, combined with expansion and compression lines
representing changes of condition of the fluid isolated in the cylinder. These
expansion and compression lines represent strictly thermal phases, laws for
which will be assumed here, but derived rigidly later in the part treating of
the thermal analysis; however, all cases can be represented by the general
expression
PV* = C,
in which the character of the case is fixed by fixing the value of s. For all
gases and for vapors that do not contain liquid or do not form or evaporate
any during expansion or compression, i.e., continually superheated, the exponent
: s may have one of two characteristic values. The first is isothermal expansion
and compression, and for this process s is the same for all substances and
equal to unity. The second is for exponential expansion or compression and
for this process s will have values peculiar to the gas or superheated vapor
under discussion, but it is possible that more than one substance may have the
Same value, as may be noted by reference to Section 8, Chapter I, from
which the value s = 1.406 for air and s = 1.3 for superheated steam or ammonia
adiabatically expanding are selected for illustration. ,
When steam or any other vapor not so highly superheated as to remain
tree from moisture during treatment is expanded or compressed in cylinders
different values of s must be used to truly represent the process and, of course,
there can be no isothermal value, since there can be no change of pressure of
wet vapors without a change of temperature. For steam expanding adiabatically
f he value of s is not a constant, as will be proved later by thermal analysis, so
that the exact solution of problems of adiabatic expansion of steam under
ordinary conditions becomes impossible by pressurevolume analysis and can
be handled only by thermal analysis. However, it is sometimes convenient
or desirable to find a solution that is approximately correct, and for this a
sort of average value for s may be taken. Rankine's average value is s = 1 . 1 1 1 = l f
jfor adiabatic expansion of ordinarily wet steam, and while other values have
ibeen suggested from time to time this is as close as any and more handy than
most. The value s = 1.035+.14X(the original dryness fraction), is given by
Perry to take account of the variation in original moisture.
Steam during expansion adiabatically, tends to make itself wet, the
Condensation being due to the lesser heat content by reason of the work done;
but if during expansion heat be added to steam originally just dry, to keep it so
continuously, as the expansion proceeds, it may be said to follow the saturation
law of steam, for which s = 1.0646. This is a strictly experimental value found
194
ENGINEERING THERMODYNAMICS
r
s
s^_
Exp.
\r
=1
Ex
,.
A
B
^s
"I
<\
I
^
^,
Exp.
Exp
Exp.
1
r

I
.
t'
X
rS=
S^E
tp.
8 =
1
I
8 =
J
5
^i
s Comp.
^c<
mp.
1
[
r
CI
j=l
P
=1
Exp.
1
sl
I
s
!
\
Exp.
\
.*Coi
pp>
1
V
V
^
 ^.
~~ _"
,c
c
M
Com]
f
f
\
\rX
Exp.
\
V
N
'
"1
\
\X
L
8
x^
Com]
)
1
Ck
mp.
Jl
= 
te
Exp.
FIG. 55. Standard Reference Cycles or Pressurevolume Diagrams for Expansive Fluid
in Simple Engines.
WORK OF PISTON ENGINES
195
by studying the volume occupied by a pound of just dry steam at various
pressures, quite independent of engines.
Direct observation of steam engine indicator cards has revealed the fact
that while, in general, the pressure falls faster at the beginning of expansion
and slower at the end than would be the case if s = l, yet the total work is
about the same as if s had this value all along the curve. This law of expansion
and compression, which may be conveniently designated as the logarithmic law,
Js almost universally accepted as representing about what happens in actual
fetaam engine cylinders. Later, the thermal analysis will show a variation of wet
ness corresponding to s = l, which is based on no thermal hypothesis what
ever, but is the result of years of experience with exact cards. Curiously
p
p
a
a
V
EXE
.PV=
c
p
V
Exp.
pys=
c
e
\
b
s
\
CYCLE 1 \.
SIMPLE ENGINE LOGARITHMIC
EXPANSION.
ZERO CLEARANCE
^
**Nfc^^
e CYCLE 2 ""^^
SIMPLE ENGINE EXPONENTIAL
EXPANSION.
ZERO CLEARANCE
C
"^ >
C
d
d
a
V
a
V
V
Exp.
PV=
p
I/
V
Exp.
?vs=
3
/
Com]
).PV=
\
c
ss
E
Com
).PYS
=c X
V

I CYCLES ^\
A SIMPLE ENGINE LOGARITH
\EXPANSION AND COMPRESS!
^K WITH CLEARANCE
viicT^
ON.
V
CYCLE 4 ^^
IMPLE ENGINE EXPONENT!
EXPANSION AND COMPRESS
V WITH CLEARANCE
fr^.
c
1
ION.
**
c
\^
X^
d
d
FIG. 56. Simple Engine Reference, Cycles or PV Diagrams.
enough, this value of s is the same as results from the thermal analysis of con
stant temperature or isothermal expansion for gases, but it fails entirely to
represent the case of isothermal expansion for steam. That s = 1 for isothermal
gas expansion and actual steam cylinder expansion is a mere coincidence, a
fact not understood by the authors of many books often considered standard,
as in them it is spoken of as the isothermal curve for steam, which it most
positively is not. This discussion of the expansion or compression laws indicates
that analysis falls into two classes, first, that for which s = l, which yields a
logarithmic expression for work, and second, that for which s is greater or
less than one, which yields an exponential expression for work, and the former
will be designated as the logarithmic and the latter as the exponential laws, for
convenience.
196
ENGINEERING THERMODYNAMICS
The phases to be considered then may be summed up as far as this analysis
is concerned as:
1. Admission or exhaust, pressure constant, P = const.
2. Admission or exhaust, volume constant, F = const.
3. Expansion, PV = const., when s = l.
4. Expansion, PV* = const., when s is greater or less than 1.
5. Compression, PV = const., when s = l.'
6. Compression, PV S = const., when s is greater or less than 1.
Considering all the possible variations of phases, there may result any or
the cycles represented by Fig. 55. These cycles have the characteristics indicated
by the following table, noting the variation in the law of the expansion or
compression that may also be possible.
Cycle.
Clearance.
Expansion.
Compression.
A
B
C
D
Zero '
Zero
Zero
Zero
Zero
Little
Complete
Overexpansion
Zero
Zero
Zero
Zero
E
F
G
H
Little
Little
Little
Little
Zero
Little
Complete
Overexpansion
Zero
Zero
Zero
Zero
I
J
K
L
Little
Little
Little
Little
Zero
Little
Complete
Overexpansion
Little
Little
Little
Little
M
N
O
P
Little
Little
Little
Little
Zero
Little
Complete
Overexpansion
Complete
Complete
Complete
Complete
Q
R
S
T
Little
Little
Little
Little
Zero
Little
Complete
Overexpansion
Too much
Too much
Too much
Too much
It is not necessary, however, to derive algebraic expressions for all these
cases, since a few general expressions may be found involving all the variables
in which some of them may be given a zero value and the resulting expression
will apply to those cycles in which that variable does not appear. The result
ing cycles, Fig. 56, that is is convenient to treat are as follows:
SIMPLE ENGINE REFERENCE CYCLE OR PV DIAGRAMS
CYCLE 1. Simple Engine, Logarithmic Expansion without Clearance.
Phase (a) Constant pressure admission.
(6) Expansion PV = const, (may be absent).
(c) Constant volume equalization of pressure with exhaust (may be absent).
(d) Constant pressure exhaust.
(e) Constant (zero) volume admission equalization of pressure with supply.
WORK OF PISTON ENGINES
197
YCLE II. Simple Engine, Exponential Expansion without Clearance.
Phase (a) Constant pressure admission.
(6) Expansion P V s = const, (may be absent).
(c) Constant volume equalization of pressure with exhasut (may be absent).
(d) Constantpressure exhaust.
" (e) Constant (zero) volume admission equalization of pressure with supply.
YCLE III. Simple Engine, Logarithmic Expansion and Compression with Clearance.
Phase (a) Constant pressure admission.
" (6) Expansion PV = const, (may be absent).
(c) Constant volume equalization of pressure with exhaust (may be absent).
(d) Constant pressure exhaust.
(e) Compression PV = const, (may be absent).
(/) Constant volume admission, equalization of pressure with supply (may
be absent).
YCLE IV. Simple Engine, Exponential Expansion and Compression with Clearance.
Phase (a) Constant pressure admission.
" (6) Expansion P V s = const, (may be absent).
(c) Constant volume equalization of pressure with exhaust (may be absent).
" (d) Constant pressure exhaust.
(e) Compression PV S = const, (may be absent).
" (/) Constant admission, equalization of pressure with supply (may be
absent).
P
A
57
?n
B
(in.pr)
(rel.pr.)
(bk.pr.)
\
\
\
>
\
\
^
V ^
\
C
E
D
D
1
~^H
G
p
Work of Expansive Fluid in Single Cylinder with No Clearance.
V
Logarithmic
Expansion for Cycle I. Exponential for Cycle II.
3. Work of Expansive Fluid in Single Cylinder without Clearance. Loga
ithmic Expansion, Cycle I. Mean Effective Pressure, Horsepower and
onsumption of Simple Engine. Referring to the diagram, Fig. 57, the net
wk, whether expansion be incomplete, perfect, or excessive, is the sum of
198
ENGINEERING THERMODYNAMICS
admission and expansion work less the backpressure work, or by are
net work area, ABODE = admission work area ABFG
+ expansion work area FBCH
backpressure work area GEDH
or algebraically,

%rP*V
(230)
Work of cycle in footpounds is
P,V, (a)
P*V d (b)
. (231)
(c)
(<D
As Vd or V e represent the whole displacement, the mean effective pressure
will be obtained by dividing Eqs. (231), by V d or V e , giving,
=P.
m.e.p. = PC
(232)
Pt (<*)
P (6)
p d (d)
P^ (e)
Similarly, dividing the Eq. (232) by the volume of fluid admitted, V b , will
give the work per cubic foot, which is a good measure of economy, greatest
economy being defined by maximum work per cubic foot, which, it may be noted, is
the inverse of the compressor standard.
Work per cu.ft. supplied = P/l+loge } P d ~ (a)
\ "/ V b
=r')P, (6)
(233)
WOEK OF PISTON ENGINES
199
According to Eq. (19), Chapter I, the piston displacement in cubic feet
13 750
>er hour per I.H.P. is for z = l>7 r]n J , and this multiplied by the fraction of
yhole displacement occupied in charging the cylinder or representing admission,
fhich is
V* V,
V d r TV
ll give the cubic feet oj high pressure fluid supplied per hour per I.H.P.,
lence
:u.ft.supplied per hr.per I.H.P. = 13 > 750
TF ()
_ 13,750 P c
~(m.e.p.) X P 6 (b)
(234)
Introducing a density factor, this can be transformed to weight of fluid. If
phen #1 is the density of the fluid as supplied in pounds per cubic foot,
Lbs. fluid supplied per hr. per I.H.P. =
13,750
13,750
. . . (235)
these expressions, Eqs. (230) to (235), for the work of the cycle, the
nean effective pressure, work per cubic feet of fluid supplied, cubic feet and
xmnds of fluid supplied per hour per I.H.P., are in terms of diagram point
jonditions and must be transformed so as to read in terms of more generally
lefined quantities for convenience in solving problems. The first step is to
ntroduce quantities representing supply and back pressures and the amount
)f expansion, accordingly:
Let (in.pr.) represent the initial or supply pressure p b expressed in pounds
per square inch;
(rel.pr.) represent the release pressure p c , in pounds per square inch;
(bk.pr.) represent the back pressure pa, in pounds per square inch;
" R v represent the ratio of expansion defined as the ratio of largest to
smallest volume on the expansion line ( ^J, or f ~] which is, of
course, equal to the ratio of supply to release pressure ( ) , when
\Pc /
the logarithmic law is assumed;
D represent the displacement in cubic feet which is V d or V c when no
clearance is assumed;
200
ENGINEERING THERMODYNAMICS
Let Z represent the fraction of stroke or displacement completed up
cutoff so that ZD represents the volume Vt> admitted to the cylinde
In this case when clearance is zero, Z=^ ,
Work of the cycle in footpounds
W=l44\(in.pT.) l+lo ^ Rv (bk. P T.)]D (a) 1
L #rj J
= 144[(rel.pr.) (1 +loge Rv) (bk.pr.)JD (6) J
m.e.p. = (rel.pr.) (1 +log e R v ) (bk.pr.) (a)
(bk.pr.) (6)
= (in.pr.)zf 1 +log e )  (bk.pr.) (c)
Work per cu.ft. supplied = 144[(in.pr.) (1 floge R v )  (bk.pr.)^ F ] (a)
.(bk^rOl
JJ
1 q 7 c A i
Cu.ft. supplied per hr. per I.H.P. = ^^ ~ (a)
13,750
(m.e.p.)
Lbs. supplied per hr. per I.H.P.
13,750 Si ,,
(m.e.p.) R v (
13,750
(m.e.p.)
ZSi(b)
(236
(237
(239
(240
The indicated horsepower may be found by multiplying the work of th
cycle, Eq. (236), by the number of cycles performed per minute n and divid
ing the product by 33,000.
H^o, (241
or
THP =
229.2
(242
In any of these expressions where R v is the ratio of greatest to smallest volume
during expansion, either R P , ratio of greater to smaller pressures, or 1, the
&
WORK OF PISTON ENGINES
201
reciprocal of the cutoff, may be substituted, since the expressions apply only
to the logarithmic law, and clearance is assumed equal to zero. When
clearance is not zero; it is shown later that the cutoff as a fraction of stroke
is not the reciprocal of R P or R v .
These expressions are perfectly general, but convenience in calculation
will be served by deriving expressions for certain special cases. The first of
these is the case of no expansion at all, the second that of complete expansion
without overexpansion. This latter gives the most economical operation from
the hypothetical standpoint, because no work of expansion has been left
unaccomplished and at the same time no negative work has been introduced
by overexpansion.
k. pr.)
FIG. 58. First Special Case of Cycles I and II. Expansion = zero. Full Stroke Admission.
First Special Case. If there is no expansion, together with the above assump
tion of no clearance, the diagram takes the form (Fig. 58), and
TP=144D[(in.pr.) (bk.pr.)] . .'
m.e.p. = (in.pr.) (bk.pr.). . . .
Work per cu.ft. supplied = 144[(in.pr.) (bk.pr.)]. . .
Cu.ft. supplied per hr. per I.H.P. =r \' /ui \
(in.pr.) (bk.pr.)
13,750Bi
Lbs. supplied per hr. per I.H.P. = ^ ^7^^. 
...... (243)
. . . . . (244)
. . . . . (245)
(246)
(247)
202
ENGINEERING THERMODYNAMICS
Second Special Case. When the expansion is complete without overexpan
sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) = (bk.pr.),
hence R v =Rp= /V?' pr 4==~ This value of cutoff, Z, is known as best cutoff, as
(bk.pr.)
U is that which uses all the available energy of the fluid by expansion.
(m.e.p.) = (in.pr.)
Work per cu.ft. supplied ==144(in.pr.)
loge Ri
R P
(248)
(249)
(250)
D (rel.pr.)=
' (bk.pr)
If .
FIG. 59. Second Special Case of Cycles I and II. Complete Expansion Without Over
Expansion Case of Best Cutoff.
Cu.ft. supplied per hr. per I.H.P.= 13 ' 7 1 50  .
(in.pr.) loge R P
Lbs. supplied per hr. per I.H.P. = ~
(in.pr.) loge R
(251)
(252)
Example 1. Method of calculating diagrams. Fig. 57 and Fig. 59.
Assumed data for Fig. 57.
Pa=P u = 90 Ibs. per sq.in. abs. V a = V e = cu.ft.
Pa =P = 14 Ibs. per sq.in. abs. V c = V d = 13.5 cu.ft.
F ft = 6cu.ft.
WORK OF PISTON ENGINES 203
To obtain point C:
y b g
PC =Pb X =90 X =40 Ibs. per sq.in. abs.
V c lo.O
Assumed data for Fig. 59.
P a =Pb= 90 Ibs. per sq.in. V a = V e = cu.f t.
P d =Pe = 14 Ibs. per sq.in. V d = 13.5 cu.ft.
To obtain point B:
^ = 13.5 X =2.1 cu.ft.
JT6 90
Example 2. A simple doubleacting engine admits steam at 100 per square inch
absolute for \ stroke, allows it to expand to the end of the stroke and then exhausts it
against a back pressure of 5 Ibs. per square inch absolute. If the engine has no
learance, a 7 X9in. cylinder and runs at 300 R.P.M., what is the horsepower and steam
onsumption when steam is expanding according to the logarithmic law? Note:
cu.ft. steam at 100 Ibs. per square inch absolute weighs .2258 Ib.
From Eq. (237 6 ),
5=54.7:
4 4
_ (m.e.p.)Lan 54.7 X. 75x38.5x600
33,000 33,000
directly from Eq. (242)
I H P =
229.2
.2X600X54.7
229.2
Lbs. steam per I.H.P. =   X ,
m.e.p. R v
=28,
=1413
Therefore, steam per hour used by engine = 14.15x28 =396 Ibs.
Prob. 1. A steam engine has a cylinder 12x18 ins. with no clearance. It runs at
200 R.P.M. and is doubleacting. If the steam pressure be fixed at 100 Ibs. per
square inch absolute, and the back pressure at 10 Ibs. per square inch abs., show how
the horsepower and steam consumption will vary as cutoff increases. Take cutoff
from i to f by eighths. Plot.
204 ENGINEEEING THERMODYNAMICS
Prob. 2. Two engines of the same size and design as above are running on a steam
pressure of 100 Ibs. per square inch absolute, but one exhausts through a long pipe to
the atmosphere, the total back pressure being 20 Ibs. per square inch absolute,
while the other exhausts into a condenser in which the pressure is but 3 Ibs. per
square inch absolute. If the cutoff is in each case f , how will the I.H.P. and steam
used in the two cases vary?
Prob. 3. By finding the water rate and the horsepower in the two following cases,
show the saving in steam and loss in power due to using steam expansively. A pump
having a cylinder 9 X 12 ins. admits steam full stroke, while an engine of same size admits
it but i of the stroke; both run at the same speed and have the same back pressure.
Prob. 4. Steam from a 12 X24 in. cylinder is exhausted at atmospheric pressure
(15 Ibs. per square inch absolute) into a tank, from which a second engine takes steam.
Neither engine has clearance. The first engine receives steam at 100 Ibs. per square
inch absolute and the cutoff is such as to give complete expansion. The second engine
exhausts into a 24 in. vacuum and its cutoff is such that complete expansion occurs in
its cylinder. Also the cylinder volume up to cutoff equals that of the first cylinder
at exhaust. If the stroke is the same in both engines and the speed of each is 200
R.P.M., what is the diameter of the larger cylinder, the total horsepower developed,
the total steam used, and the work per cubic foot of steam admitted to the first
cylinder, the water rate of each engine and the total horsepower derived from each
pound of steam?
Prob. 5. The steam pressure for a given engine is changed from 80 Ibs. per square
inch gage to 120 Ibs. per square inch gage. If the engine is 12x16 ins., running 250
R.P.M. with a fixed cutoff of 25 per cent and no clearance, the back pressure being
15 Ibs. per square inch absolute, what will be the horsepower and the water rate in
each case?
NOTE: 1 cu.ft. of steam at 80 and 120 Ibs. weighs .215 and .3 Ib. respectively.
Prob. 6. By trial, find how much the cutoff should have been shortened to
keep the H.P. constant when the pressure was increased and what effect this would
have had on the water rate.
Prob. 7. A certain type of automobile engine uses steam at 600 Ibs. per square
inch absolute pressure. The exhaust is to atmosphere. For a cutoff of  and no
clearance, what would be the water rate?
NOTE: for 600 Ibs. <?i=1.32.
Prob. 8. Engines are governed by throttling the initial pressure or shortening the
cutoff. The following cases show the effect of light load on economy. Both engines,
12x18 ins., running at 200 R.P.M., with 125 Ibs. per square inch absolute. Initial
pressure and back pressure of 10 Ibs. per square inch absolute. The load is sufficient
to require full steam pressure at \ cutoff for each engine. Load drops to a point
where the throttle engine requires but 50 Ibs. per square inch absolute initial pressure
with the cutoff still fixed at . What is the original load and water rate, and new
load and waterrate for each engine?
NOTE: a for 125 Ibs. absolute = .279 and for 50 Ibs. =.117 Ib.
Prob. 9. The guarantee for a simple engine 18x24 ins., running at 200 R.P.M.,
states that the I.W.R. when cutoff is } will not exceed 15 Ibs. if the initial pressure be
100 Ibs. per square inch gage, and back pressure 5 Ibs. per square inch absolute.
If engine has no clearance, see if this would be possible.
4. Work of Expansive Fluid in Single Cylinder without Clearance. Expo
nential Expansion Cycle II. Mean Effective Pressure, Horsepower and
WORK OF PISTON ENGINES
205
Consumption of Simple Engines. Referring to the diagram, Fig. 57, the
work is given by the same areas as for Cycle I, Hut its algebraic expression
is different because s is greater than 1 and an exponential expansion results on
integration instead of a logarithmic one.
In general, from Eq. (13a), Section 7, Chapter I,
v
Putting this in terms of initial conditions by the relations
/PA /V c \* *
\^p) = (~v) = #F and F < =F <* =
\* / \ ' b/
Also
there results
W = P*V*+, _ P *^ b 1 (Rv s  l l)P d V^
,
{S
 1 i) 
= 144D \Z (in.pr.) ^^p  (bk.pr.)l
(c)
which is the general equation for work of this cycle.
Dividing by 7 6 , the volume of fluid supplied,
Work per cu.ft. supplied = .
_ 1 P d Rv (a)
(253)
(254)
(255)
Similarly, the mean effective pressure results from dividing the work by the
displacement, V d =Vi,Rv
or
(a)
Jtbk.pr.) (6)
. (256)
(c)
206
ENGINEERING THERMODYNAMICS
First special case of no expansion, when /iV = l> results the same diagran
as in the previous section, Fig, 58, and exactly the same set of formulas.
Second special case when the expansion is complete without overexpansion
is again represented by Fig. 59 and for it
Whence
or
Work for complete expansion is
R
(6)
()
. . (257
(D \
which is the general equation for the work of V b or ( ^r ) cubic feet of fluic
when the economy is best or for best cutoff.
The work per cubic foot of fluid supplied for this case of complete expansioi
gives the maximum value for Eq. (255) and is obtained by dividing Eq. (257
Max. work per cu.ft. supplied
, .(258
which is the general equation for maximum work per cubic foot of fluid supplied
The expression for mean effective pressure becomes for this case of best cutoff,
or,
. . . (259
It is convenient to note that in using Eqs. (257), (258) and (259) it may b
desirable to evaluate them without first finding R v . Since
Vc
v
P.
=p
WORK OF PISTON ENGINES 207
}his substitution may be made, and
sl
Z? 7">
fly lp *
(Example. Compare the horsepower and the steam consumption of a 9x12 in.
simple doubleacting engine with no clearance and running at 250 R.P.M. when initial
pressure is 100 Ibs. per square inch absolute and cutoff is J, if
(a) steam remains dry and saturated throughout expansion,
(6) remains superheated throughout expansion, and
(c) if originally dry and suffers adiabatic expansion.
Back pressure is 10 Ibs. per square inch absolute.
For case (a) . 1.0646 and (m.e.p.) ~ .w 10 =48.6;
" (6) S = 1.3 and (m.e.p.) J?(!_L) 10=43.5;
(c). 1.111 and fr*PJT(lir.iilx4m) 10=47.5.
/. I.H.P. for case (a) =46.9,
(6) =42.0,
(c) =45.8.
From Eq. (240), Ibs. steam per hour per I.H.P. =.X
m.e.p.
"I Q *7 PCn 5s
.*. For case (a) steam per hr. =46.9 X ' . XT 5
i 48.o 4
" (6) steam per hr. =42 X X j;
" (c) steam per hr. =45.8 Xi X.
47.5 4
Prob. 1. On starting a locomotive steam is admitted full stroke, while in running
he valve gear is arranged for f cutoff. If the engine were 18x30 ins., initial pressure
150 Ibs. per square inch absolute, back pressure 15 Ibs. per square inch absolute, what
would be the difference in horsepower with the gear in normal running position and
in the starting position for a speed of 20 miles per hour with 6ft. driving wheels? Con
sider the steam to be originally dry and expanding adiabatically. What would be the
difference in steam per horsepower hour for the two cases and the difference in total
steam? Clearance neglected.
Prob. 2. Consider a boiler horsepower to be 30 Ibs. of steam per hour; what must be
the horsepower of a boiler to supply the following engine? Steam is supplied in a super
208 ENGINEERING THERMODYNAMICS
heated state and remains so throughout expansion. Initial density of steam = .21 Ibs. per
cubic foot. Engine is 12x20 ins., doubleacting, 200 R.P.M., no clearance, initial
pressure 125 Ibs. per square inch absolute, back pressure a vacuum of 26 ins. of
mercury. Cutoff at maximum load f, no load, f&. What per cent of rating of boiler
will be required by the engine at no load?
Prob. 3. While an engine driving a generator is running, a short circuit occurs
putting full load on engine, requiring a f cutoff. A moment later the circuitbreaker
opens and only the friction load remains, requiring a cutoff of only . The engine
being twocylinder, doubleacting, simple, 12 X18 ins., running at 300 R.P.M., and having
no clearance, what will be the rate at which it uses steam just before and just after
circuitbreaker opens if the steam supplied is at 125 Ibs. per square inch absolute and is
just dry, becoming wet on expanding, and back pressure is 3 Ibs. per square inch
absolute?
Prob. 4. A pumping engine has two doubleacting steam cylinders each 9X12 ins.
and a fixed cutoff of \. It runs at 60 R.P.M. on 80 Ibs. per square inch absolute steam
pressure and atmospheric exhaust. Cylinder is jacketed so that steam stays dry
throughout its expansion. How much steam will it use per hour? Neglect clearance.
Prob. 6. If an engine 10x14 ins. and running 250 R.P.M. has such a cutoff that
complete expansion occurs for 90 Ibs. per square inch absolute initial pressure, and at
atmospheric (15 Ibs. absolute) exhaust, what will be the horsepower and steam used per
hour, steam being superheated at all times, and what would be the value for the horse
power and steam used if full stroke admission occurred?
Prob. 6. The steam consumption of an engine working under constant load is
better than that of a similar one working under variable load. For a 16 X 24 ins. engine
running at 250 R.P.M. on wet steam of 125 Ibs. per square inch absolute and atmospheric
exhaust, find the horsepower and steam used per horsepower per hour for best con
dition and by taking two lighter and three heavier loads, show by a curve how steam
used per horsepower per hour will vary.
Prob. 7. For driving a shop a two cylinder singleacting engine, 6x6 ins., running
at 430 R.P.M., is used. The cutoff is fixed at \ and intitial pressure varied to control
speed. Plot a curve between horsepower and weight of steam per hour per horse
power for 20, 40, 60, 80, 100, 120 Ibs. per square inch absolute initial pressure and
atmospheric exhaust. Steam constantly dry. Clearance zero.
NOTE: <*i for above pressures equals .05, .095, .139, .183, .226, and .268 Ibs. per
cubic foot, respectively.
Prob. 8. Taking the loads found in Prob. 7, find what cutoff would be required j
to cause the engine to run at rated speed for each load if the initial steam pressure
were 100 Ibs. per square inch absolute, and the back pressure atmosphere, and a plot
curve between horsepower and steam used per horsepower hour for this case.
Prob. 9. For working a mine hoist a twocylinder, doubleacting engine is used
in which compressed air is admitted f stroke at 125 Ibs. per square inch absolute and then
allowed to expand adiabatically and exhaust to atmosphere. If the cylinders are 18 X24
ins. and speed is 150 R.P.M., find the horsepower and cubic feet of high pressure air
needed per minute.
5. Work of Expansive Fluid in Single Cylinder with Clearance. Loga
rithmic Expansion and Compression; Cycle III. Mean Effective Pressure,
Horsepower, and Consumption of Simple Engines. As in previous cycles,
the net work of the cycle is equal to the algebraic sum of the positive work
WORK OF PISTON ENGINES
209
one on the forward stroke and the negative work on the return stroke. By
Teas, Fig. 60, this is
Work area = JABN+NBCW WDEO  OEFJ.
Expressed in terms of diagram points this becomes
W
(260)
(CD)
c
upply
Yolun
7R 
e
,
B
in.pr.
"V
ft
Q
\
i
\
i
X
\
F
\
^
"^
^
^C(t
\
\
\
.__
..
\
D(
Ib
1
1
"M
W
j
!c
N
D
KXD^I V
l.pr)
k.pr)
. 60. Work of Expansive Fluid in Single Cylinder with Clearance. Logarithmic Expan
sion for Cycle III. Exponential for Cycle IV.
Expressing this in terms of displacement, in cubic feet D; clearance as a frac
iion of displacement, c; cutoff as a fraction of displacement, Z; compres
sion as a fraction of displacement, X; initial pressure, in pounds per square
nch (in.pr.), and exhaust or back pressure, in pounds per square inch (bk.pr):
P 6 = 144(in.pr.);
P d = 144 (bk.pr.);
(Vt > V a )=ZD' )
D(Z+c) Z+c'
V e = D(X+c).
V e = D(X+c) = X+c
V f DC c '
210
Whence
Work in ft.lbs. per cycle is
144D j (in.pr.)
W
(bk.pr.) 1(1
ENGINEERING THERMODYNAMICS
) l oge
(261)
From Eq. (261), the mean effective pressure, pounds per square inch, follows
by dividing by 144D:
(in.pr.)
 (bk.pr.) (l
. . . (mean forward press.)
) loge ^~t c .J (
(262)
mean bk.pr.)
This is a general expression of very broad use in computing probable mean
effective pressure for the steam engine with clearance and compression, or for
other cases where it is practicable to assume the logarithmic law to hold. Fig.
117, at the end of this chapter, will be found of assistance in evaluating this
expression.
Indicated horsepower, according to expressions already given, may be
found by either of the following equations:
33,000
144(m.e.p.)Dn = (m.e.p.)Dn
33,000 229.2~~'
where L is stroke in feet, a is effective area of piston, square inches, n is the
number of cycles performed per minute and D the displacement, cubic feet.
It might seem that the work per cubic foot of fluid supplied could be
found by dividing Eq. (261) by the admission volume,
but this would be true only when no steam is needed to build up the pressure
from F to A. This is the case only when the clearance is zero or when com
pression begins soon enough to carry the point F up to point A, i.e., when by
compression the pressure of the clearance fluid is raised to the initial pressure.
It is evident that the fluid supplied may perform the two duties: first,
building up the clearance pressure at constant or nearly constant volume,
and second, filling the cylinder up to cutoff at constant pressure. To measure
the steam supplied in terms of diagram quantities requires the fixing of the
volume of live steam necessary to build up the pressure from F to A and adding
it to the apparent admission volume (V b V a ). This can be done by producing
WOEK OF PISTON ENGINES 211
,he compression line EF to the initial pressure Q, then LQ is the volume that
ihe clearance steam would have at the initial pressure and QA the volume of
ive steam necessary to build up the pressure. The whole volume of steam
tdmitted then is represented by QB instead of AB or by (V b V g ) instead of
^y (Vb Va), and calling this the supply volume,
But
Hence
(Sup.Vol.) =D\(Z\c) (X\c)~ Jy , .... (263)
which is the cubic feet of fluid admitted at the initial pressure for the dis
placement of D cubic feet by the piston. Dividing by D there results
(264)
which is the ratio of admission volume to displacement or cubic feet of live
steam admitted per cubic foot of displacement.
Dividing the work done by the cubic feet of steam supplied gives the
economy of the simple engine in terms of volumes, or
W
Work per cu.ft. of fluid supplied =
(Sup.Vol.)
(in.pr.) \Z+ (Z+c)log e J^~  (bk.pr.) ["(1 X) + (X+c)lo&
' (265)
It is more common to express economy of the engine in terms of the weight
of steam used per hour per horsepower or the " water rate," which in more
general terms may be called the consumption per hour per I.H.P.
Let di be the density or weight per cubic foot of fluid supplied, then the
weight per cycle is (Sup.Vol.) di, and this weight is capable of performing
W
W footpounds of work or (Sup.Vol.) di Ibs.. per minute will permit of
horsepower. But (Sup.Vol.) 3i Ibs. per mintue corresponds to 60 (Sup.Vol.)
L Ibs. per hour, whence the number of pounds per hour per horsepower is
60 (Sup.Vol.)fli
TF/33,000 '
212 ENGINEERING THERMODYNAMICS
which is the pounds consumption per hour per I.H.P., whence
60X33,000(Sup.Vol.)3i
Consumption in Ibs. per hr. per I.H.P.= ~~W~ *''' ' * ' '
which is the general expression for consumption in terms of the cubic feet of
fluid admitted per cycle, ^i initial density, and the work per cycle.
As work is the product of mean effective pressure in pounds per square foot,
(M.E.P.,) and the displacement in cu.ft. or W = (M.E.P.)D, or in terms of
mean effective pressure pounds per square inch TF = 144 (m.e.p.)D, the
consumption may also be written
60X33,000(Sup.Vol.)3i
Consumption in lbs.,per hr. per I.H.P. = i44( m . e .p.)D
13,750 (Sup.Vol.)3i
(m.e.p.) D
13,750 , , y . v(bk.pr.)1
 * lt ..... (267)
which gives the water rate in terms of the mean effective pressure, cutoff,
clearance, compression, initial and back pressures and initial steam density.
It is sometimes more convenient to introduce the density of fluid at the back
pressure ^2, which can be done by the relation (referring to the diagram),
p v _ PF or (in.pr.) _V e _di
M^ ^ K ' 0r (bk.pr:) ^ S
whence
(bk.pr.)
This on substitution gives
Consumption in Ibs., per hr. per I.H.P.
Since the step taken above of introducing ^2 has removed all pressure or
volume ratios from the expression, Eq. (268) is general, and not dependent
upon the logarithmic law. It gives the consumption in terms of mean effective
pressure, cutoff, clearance, compression and the density of steam at initial
and back pressure, which is of very common use.
It cannot be too strongly kept in mind that all the preceding is true only when
no steam forms from moisture water during expansion or compression or no steam
condenses, which assumption is known to be untrue. These formulae are, there
fore, to be considered as merely convenient approximations, although they
WORK OF PISTON ENGINES
213
ire almost universally used in daily practice. (See the end of this chapter for
liagrams by which the solution of this expression is facilitated.;
Special Cases. First, no expansion and no compression would result in
Fig. 61. For it
TF=144Z)[(in.pr.) (bk.pr.)] (269)
(m.e.p.) = (in.pr.)(bk.pr.) . . (270)
p
L
n
Su]
plyV
ft . 1T _
_,
B
(in.pr.)
(bk.pr.)
j
JQ
cD
i
i
N
\
D
i
K
p
w v
FIG. 61. First Special Case of Cycles III and IV. Expansion and Compression both Zero,
but Clearance Finite.
The volume of fluid supplied per cycle is QB, or from Eq. (263) it is
(271)
13,750 f (bk.pr.)l,
Consumption in Ibs. per hr. per I.H.P.= ^  N ,, ,  , 1 +cc r ~ \oi
(m.pr.)  (bk.pr.) _ (in.pr.) ] 2
or in terms of initial and final densities,
13 750
'
(272)
Consumption in Ibs. per hr. per l.H.P. = T .
(273)
The second special case is that of complete expansion and compression, as
indicated in Fig. 62. Complete expansion provides that the pressure at the
214
ENGINEERING THERMODYNAMICS
end of expansion be equal to the back pressure, and complete compression that
the final compression pressure be equal to the initial pressure.
Here
and hence
and
_ = _ = X+c = (in.pr.)
F 6 Fa Z+c ' c (bkpr.)'
F c F_in.pr. y \p\i I c c (' m ' w '\\
V^Fa'bk^F.' ^L \bk.prJJ
(V b Va)=ZD.
in.pr.)
a~ (bk.pr.)
FIG. 62. Second Special Case of Cycles III and IV. Perfect Expansion and Perfect
Compression with Clearance.
Hence by substitution
ZD
(bk.pr.) '
from which
Again,
v.v t
D
[(&H

WORK OF PISTON ENGINES 215
5q. (274) gives the cutoff as a fraction of the displacement necessary to give
complete expansion, while Eq. (275) gives the compression as a fraction of dis
jlacement to give complete compression, both in terms of clearance, initial
Dressure and back pressure, provided the logarithmic law applies to expansion
and compression.
Substitution of the values given above in Eq. (261) gives, after simplifi
3ation,
TF = 144D[(l+c)(bk.pr.)c(in.pr.)]log e ^^ . . (276)
..... (277)
sin this case the volume supplied is exactly equal to that represented by the
admission line AB, and is equal to
(Sup.Vol.)=ZD. . . . (278)
Hence, the consumption, in pounds fluid per hour per I.H.P. in terms of
initial density, is
13 750
Consumption in Ibs. per hr. per I.H.P. = ' r Zd\ t
(m.e.p.)
but
Z
m.e.p. ,. . f_ , .(bk.pr.) 1, (in.pr.) ' X1 (in.pr.)
(in.pr.) (\\c). c log e /U1 N (in.pr.) logeTrr^ :
' [ (m.pr.) J 3 (bk.pr.) & e (bu^>
hence
13 ' tyv^ i
Consumption in Ibs. fluid per hr. per I.H.P. = ^^ r. . . (279)
/ \ i (m.pr.)
(m.pr.) ^ v
This last equation is interesting in that it shows the consumption (or water
rate, if it is a steam engine) is independent of clearance, and dependent only
upon initial density, and on the initial and final pressures.
An expression may also be easily derived for the consumption in terms of
initial and final density, but due to its limited use, will not be introduced here.
Example 1. Method of calculating Diagrams Fig. 60 and Fig. 62.
Assumed data for Fig. 60:
p a =p a =p b = 90 Ibs. per square inch abs. V a = V f = .5 cu.ft.
P g =p e =P d = 14 Ibs. per square inch abs. V d = V c = 135 cu.ft.
P f = 50 Ibs. per square inch abs. F& = 6 cu.ft.
216 ENGINEERING THERMODYNAMICS
To obtain point C:
p t =p b X =90X =40 Ibs. per sq.in.
To obtain point E:
7 e = F/X / = .5x = 1.78cu.ft.
Pe 14
To obtain point Q:
P f 50
Intermediate points from B to C and E to Q are found by assuming volumes
and computing the corresponding pressures by relation P x V x =PbVb or P x V x =P e V e .
Clearance is  A=r =~ =3.8 per cent,
Vd V a 16
Cutoff is ft ~ a =^ =42.3 per cent,
F V a 1 28
Compression is = =^r =9.9 per cent.
V d~ Va lo
Assumed data for Fig. 62.
p a =p 6 =90 Ibs. per square inch absolute. V a = .5 cu.ft. ,
P e = Pc = 14 Ibs. per square inch absolute. Vd = 13.5 cu.ft.
To obtain point B:
V b = V C ^= 13.5 X^: =2.11 cu.ft.
To obtain point E:
rr TT . .*a *J1J
Intermediate points from B to C and from A to E are to be found by assuming
various volumes and finding the corresponding pressures from relation P x V x =PaVa or
p t v x =p b v b .
Example. 2. What will be the horsepower of, and steam used per hour by the
following engine:
(a) cutoff 50 per cent, compression 30 per cent,
(6) complete expansion and compression,
(c) no expansion or compression.
Cylinder, 12 Xl8in. doubleacting, 200 R.P.M., 7 per cent clearance, initial pressure
85 Ibs. per square inch gage, back pressure, 15 Ibs. per square inch absolute, and
logarithmic expansion and compression.
WORK OF PISTON ENGINES 217
NOTE: d for 85 Ibs. gage = .23, <?i for 15 Ibs. absolute = .038 cu.ft.
(a) From Eq. (262)
(bk.pr.) (1 
log e 
=86 20 =66 Ibs. sq.in.
^ (m.e.p.) Lan ^66X1.5X113.1X400
33,000 33,000
From Eq. (267) steam per hour per I.H.P. in pounds is
13,750 [., . /bk.prAl
,  r\(Z+c)(X+c)\T 8l,
(m.e.p.) _ V 'Vm.pr./J '
Hence total steam per hour =25x1 35 =3380 Ibs.
(6) From Eq. (277)
(m.e.p.) =[(1 +c)(bk. pr.) c(in. pr.)] lo ge ,
=[(1+.07) X15.07X100] log e 6.67 = 17.2 Ibs. sq.in.
17.2X1.5X113.1X400
LH * R= ~^3,00^~
From Eq. (279)
13,750 Si 13,750 X. 23
Steam per I.H.P. per hour =  /V= Too~xl^
Total steam per hour = 16.6x35. 4 =588 Ibs.
(c) From Eq. (270) (m.e.p.) =(in.pr.) (bk.pr.) =10015=85 Ibs. sq.in.
Eq. (273)
1 3 7^0 1 ^ 7^0
Steam per I.H.P. per hour = ,^[(1 +c)*i o 2 ]  ^[1.07x.23 .07X.038] =35.4.
(m.e.p.) 85
Total steam per hour = 174.5x35.4 =6100 Ibs.
Prob. 1. What will be the horsepower and water rate of a 9x12 in. simple engine
having 5 per cent clearance, running at 250 R.P.M. on 100 Ibs. per square inch abso
218 ENGINEERING THERMODYNAMICS
lute initial pressure and 5 Ibs. per square inch absolute back pressure when the cutoff is
, J, and , expansion follows the. logarithmic law, and there is no compression?
NOTE: 8 for 100 Ibs. absolute = .23, 8 for 5 Ibs. absolute = .014.
Prob. 2. Will a pump with a cylinder 10x15 ins. and 10 per cent clearance give the
same horsepower and have the same water rate as a pump with cylinders of the same
size but with 20 per cent clearance, both taking steam full stroke? Solve for a case
of 125 Ibs. per square inch absolute initial pressure, atmospheric exhaust and a speed of
50 double strokes. No compression.
NOTE: 8 for 125 Ibs. absolute = .283, 8 for 15 Ibs. absolute = .039.
Prob. 3. Solve the above problem for an engine of the same size, using steam expan
sively when the cutoff is \ and R.P.M. 200, steam and exhaust pressure as in Prob. 2
and compression zero.
Prob. 4. Two engines, each 9x15 ins., are running on an initial pressure of 90 Ibs.
per squareHnch absolute, and a back pressure of one atmosphere. One has no clearance,
the other 8 per cent. Each is cutting off so that complete expansion occurs. The speed
of each is 200 and neither has any compression. What will be the horsepower and
water rate?
NOTE: 8 for 90 Ibs. =.24, 8 for 15 Ibs. = .039.
Prob. 5. By finding the horsepower and water rate of a 12x18 in. doubleacting
engine with 8 per cent clearance, running at 150 R.P.M. on an initial steam pressure
of 90 Ibs. per square inch absolute and atmosphere exhaust for a fixed cutoff of \ and
variable compression from to the point where the pressure at the end of compression
is equal to 125 per cent of the initial pressure, plot the curves between compression and
horsepower, and compression and water rate to show the effect of compression on
the other two.
NOTE: 8 for 90 Ibs. =.21, 8 for 15 Ibs. =.039.
Prob. 6. A steam engine is running at such a load that the cu^off has to be f at
a speed of 150 R.P.M. The engine is 14 X20 ins. and has no clearance. Initial pressure
100 Ibs. per square inch absolute and back pressure 5 Ibs. per square inch absolute.
What would be the cutoff of an engine of the same dimensions but with 10 per cent
clearance under similar conditions?
Prob. 7. The steam pressure is 100 Ibs. per square inch gage and the back
pressure is 26 ins. of mercury vacuum when the barometer is 30 ins. For a 16x22
in.engine with 6 per cent clearance running at 125 R.P.M., cutoff at \ and 30 per cent
compression, what will be the horsepower and the water rate? Should the steam
pressure be doubled what would be the horsepower and the water rate? If it should
be halved?
NOTE: 8 for 100 Ibs. gage = .2017, 8 for 26 ins. Hg.=.0058.
Prob. 8. While an 18x24 in. simple engine with 4 per cent clearance at speed
of 150 R.P.M. is running with a \ cutoff and a compression of \ on a steam pressure of
125 Ibs. per square inch gage, and a vacuum of 28 ins. of mercury, the condenser fails
and the back pressure rises to 17 Ibs. per square inch absolute. What will be the change in
the horsepower and water rate if all other factors stay constant? What would the new
cutoff have to be to keep the engine running at the same horsepower and what
would be the water rate with this cutoff?
NOTE: 8 for 125 Ibs. gage =.315, 8 for 28 in. Hg.=.0029, 8 for 17 Ibs. absolute
= .0435.
Prob. 9. Under normal load an engine has a cutoff of , while under light load
the cutoff is but A. What per cent of the steam used at normal load will be used
WOEK OF PISTON ENGINES 219
it light load for the following data? Cylinder 10x18 ins.; 7 per cent clearance; 200
R.P.M.; initial pressure 120 Ibs. per square inch gage; back pressure 2 Ibs. per square
nch absolute; compression at normal load 5 per cent; at light load 25 per cent.
NOTE: S for 120 Ibs. gage =.304, 8 for 2 Ibs. absolute = .0058.
II
6. Work of Expansive Fluid in Single Cylinder with Clearance ; Exponential
Expansion and Compression, Cycle IV. Mean Effective Pressure, Horse
)ower and Consumption of Simple Engines. As pointed out in several places,
;he logarithmic expansion of steam only approximates the truth in real engines
rnd is the result of no particular logical or physically definable hypothesis as to
}he condition of the fluid, moreover its equations are of little or no value
r or compressed air or other gases used in engine cylinders. All expansions
jjvhat can be defined by conditions of physical state or condition of heat,
Including the adiabatic, are expressible approximately or exactly by a definite
/alue of s, not unity, in the expression PV S = const. All these cases can
hhen be treated in a group and expressions for work and mean effective
3ressure found for a general value of s, for which particular values belonging
x), or following from any physical hypothesis can be substituted. The area
mder such expansion curves is given by Eq. (13) Chapter I, which applied to
I:he work diagram, Fig. 60, in the same manner as was done for logarithmic
:j 3xpansion, gives the net work:
sl
(area JABCWJ) _
. (280)
(area WDEFJW )
~~ L L \ v // J
Introducing the symbols,
P 6 = 144(in.pr.), V b = D(Z+c).
P d = 144(bk.pr.), V e = D(X+c).
(VV a )=ZD, (T.rm
(Vd VJ=D(lX), =?'
. . (281)
Eq. (281) gives the work in footpounds for D cubic feet of displacement in a
jylinder having any clearance c, cutoff Z, and compression X, between two
220 ENGINEERING THERMODYNAMICS
pressures, and when the law of expansion is PV S = const, and s anything except
unity, but constant.
The mean effective pressure, pounds per square inch, is obtained by divid
ing the expression for work by 144Z), giving
, (282)
(m.c.p.) = (in.pr.) j Z+^j[l  (5)" ( mean for>d
(bk.pr.) ( (l_JE)+gK(*?y Vi] 1 (mean bk.pr.)
[ s ~ L L \ c / J J
which is the general expression for mean effective pressure for this cycle.
It was pointed out in Section (5) that the cubic feet of fluid admitted at the
initial pressure was not represented by AB, Fig. 60, but by QB, and the same
is true for this case, so that the
But when the expansion and compression laws have the form PV s =
Whence
^y ..... (283)
Eq. (283) gives for these cases what was given by Eq. (263) for the logarithmic
law, the cubic feet of fluid supplied at the initial pressure for the displacement
of D cubic feet in terms of cutoff, clearance, compression and the pressures.
From this, by division there is found the volume of high pressure fluid per
cu.ft. of displacement
(Sup.Vol.)
D
The consumption is given by the general expression already derived,
Eq. (34), from which is obtained,
Consumption Ibs. per hr. per I.H.P.
WORK OF PISTON ENGINES 221
Eq. (285) gives the water rate or gas consumption in terms of mean effective
pressure, initial and back pressure, cutoff, clearance, compression and initial
fluid density. Introducing the density at the back pressure by the relation,
PIT _ p V s
a V Q * e r e )
J^
jj = Si /in.pr.X *
V, 8 2 Vbk.pr./
i.pr.AT
k^rV '
in.
bk.
[there results
Consumption Ibs. per hr. per I.H.P. = ^J 5 \(Z+c) 81  (X+c) 8 2 1 , .
(286)
rhich is identical with Eq. (268) and is, as previously observed, a general expres
sion, no matter what the laws of expansion and compression, in terms of
mean effective pressure, cutoff, clearance, compression and the initial and
final steam density.
The first special case of full admission, no compression might at first thought
appear to be the same as in the preceding section, where the logarithmic law
was assumed to hold, and so it is as regards work and mean effective pressure,
Eq. (269) and (270), but referring to Fig. 61 it will be seen that since the expo
nential law is now assumed instead of the logarithmic, the point Q will be dif
ferently located (nearer to A than it was previously if s is greater than 1), and
hence the supply volume QB is changed, and its new value is
in.pr.
(287)
Hence, consumption stated in terms of initial density of the fluid 81, is
Consumption Ibs. per hr. per I.H.P.
13 ' 75
(in.pr.)(bk.pr.)
lie
. . (288)
Stated in terms of initial and final densities, the expression is as before, Eq. (273).
The second special case, complete expansion and compression is again repre
sented by Fig. 62. From the law of expansion it is evident that
222 ENGINEERING THERMODYNAMICS
or stated in symbolic form,
whence
Again referring to Fig. 62,
'^ ?Y cD
V c Va
Whence
y re 'a ywt.pt.
A = ^~
(290)
Eq. (289) gives the cutoff as a fraction of displacement necessary to give
complete expansion, and (290), the compression fraction to give complete com
pression, both in terms of clearance, initial and back pressures, and the exponent
s, in the equation of the expansion or compression line, PV S const.
The work of the cycle becomes for this special case, by substitution in Eq.
(281),
J^ s1
F=144Z)(in.pr.)
and the mean effective pressure, Ibs. per sq.in., is
( 292 )
hence
The volume of fluid supplied is,
(Sup.Vol.)=ZD, .... .... (293)
Consumption, Ibs. per hr. per I.H.P.=
m.e.p
but
Z
81
r ^ s /bk.prA s
(m.pr.) 1  ^~
y sl[ \in.pr. / J,
WOEK OF PISTON ENGINES 223
whence
Consumption Ibs. fluid per hr. per I.H.P. is,
13,750XSi
sl
(294)
,. , s /bk.pr.X "I
(m.pr.) 1 
slL \in.pr. /
the expression for smallest consumption (or water rate if steam) of fluid for
the most economical hypothetical cycle, which may it be noticed, is again in
dependent of clearance.
The expressions for work and mean effective pressure are not, however,
independent of clearance, and hence, according to the hypothetical cycles here
considered, it is proved that large clearance decreases the work capacity of a
a cylinder of given size, bjit does not affect the economy, provided complete
expansion and compression are attained, a conclusion similar to that in regard
to clearance effect on compressor capacity and economy. Whether the actual
performance of gas or steam engines agrees with this conclusion based only
on hypothetical reasoning, will be discussed later.
Example 1. What will be the horsepower of and steam used per hour by the
following engine: 12xl8in. doubleacting, 200 R.P.M., 7 per cent clearance, initial
pressure 85 Ibs. per square inch gage, back pressure 15 Ibs. per square inch absolute,
and expansion such that s = 1.3.
(a) cutoff =50 per cent; compression = 30 per cent;
(b) complete expansion and compression;
(c) no expansion or compression;
NOTE: 8 for 85 Ibs. gage =.18; for 15 Ibs. absolute. = .03.
(a) From Eq. (282)
59.8X1.5X113.1X400

From Eq. (286)
Steam per hour per I.H.P.
,  X.18(.37) x.03] =20.9 Ibs.
(m.e.p.) _ j 5
/. Steam per hour = 122 X20.9 = 2560 Ibs.
224 ENGINEERING THERMODYNAMICS
(6) From Eq. (292)
s f x /bk.pr.\r "ir /bk.pr.\~~
(m.e. P = (in.pr.) I [(l +C )(y ^Jl^ta^ J'
sl
I .in.,
26.2X1.5X113.1X200
3000
From Eq. (294) steam used I.H.P. per hour is,
_ 13,7508i _ 13,750 X.18 _ ==16>51bs . )
T /bk.prAT 1 ] 1.3 I" /ISXTT]
(p^rTL^tpTrj J ^"T'ViooJ J
hence total steam per hour = 16.5 X64 = 1060 Ibs.
(c) From Eq. (270) which holds for any value of s, m.e.p. =10015 =85 Ibs. sq.in.
From Eq. (288) steam per I.H.P. hour
1 =24 5 lbs
(m.e.p.) L \in.pr. / J 85
and total steam per hour = 174.5X24.5 =2475 lbs.
Prob. 1. Two simple engines, each 12 Xl8 ins., with 6 per cent clearance are operated
at ^ cutoff and with 20 per cent compression. One is supplied with air at 80 lbs.
per square inch gage, and exhausts it to atmosphere; the other with initially dry steam
which becomes wet on expansion and which also is exhausted to atmosphere. For a speed
of 200 R.P.M. what is the horsepower of each engine and the cubic feet of stuff supplied
per horsepower hour ?
Prob. 2. A crankandflywheel twocylinder, doubleacting, pumping engine is
supplied with dry steam and the expansion is such that it remains dry until exhaust.
The cylinder size is 24x36 ins., cutoff to give perfect expansion, clearance 5 per cent,
compression to give perfect compression, initial pressure 50 lbs. per square inch ab
solute. back pressure 5 lbs. per square inch absolute. What is the horsepower and
water rate? What would be the horsepower and water rate of a fullstroke pump of the
same size and clearance but having no compression, running on the same pressure range
and quality of steam.
NOTE: 8 for 50 lbs. absolute = .12, 8 for 15 lbs. absolute = .0387.
Prob. 3. Should the cylinder of the following engine be so provided that th<
steam was always kept dry, would there be any change in the horsepower developed a:
WORK OF PISTON ENGINES 225
compared with steam expanded adiabatically, and how much? Cylinder 20x24 ins.,
initial pressure 125 Ibs. per sq. in. gage, back pressure 26 ins. vacuum, standard barom
eter, clearance 6 per cent, cutoff f, compression 10 percent, and speed 125 R.P.M.
Prob. 4. What will be the total'steam used per hour by a 20x28in. doubleacting
engine running at 150 R.P.M. if the initial pressure be 125 Ibs. per square inch absolute,
back pressure one atmosphere, clearance 8 per cent, compression zero, for.cutoff
I, i, f, and %, if steam expands adiabatically and is originally dry and saturated?
NOTE: 8 for 125 Ibs. absolute = .283, 8 for 15 Ibs. absolute = .0387.
Prob. 5. An engine which is supplied with superheated steam is said to have an
indicated water rate of 15 Ibs. at  cutoff and one of 25 Ibs. at  cutoff. See if
this is reason able for the folio wing conditions: engine is 15x22 ins., 7 per cent clearance,
no compression, initial pressure 100 Ibs. per square inch gage., back pressure 28in.
vacuum, barometer 30 ins. and speed 180 R.P.M.
NOTE: 8 for 100 Ibs. gage =.262, 8 for 28 in. Hg = .0029.
Prob. 6. The tank capacity of a locomotive is 4500 gals. The cylinders are
18X36 ins., initial pressure 200 Ibs. per square inch gage, exhaust atmospheric, cutoff 5,
clearance 7 per cent, speed 200 R.P.M., no compression. The stream is dry at
start and expansion adiabatic, how long will the water last if 40% condenses during
admission?
NOTE: 8 for 200 Ibs. gage = .471, 8 for 15 Ibs. absolute = .0387.
Prob. 7. To drive a hoist, an air engine is used, the air being supplied for ^
stroke at 80 Ibs. per squjfc&inch gage expanded adiabatically and exhausted to atmos
sphere. If the clearance is Sjf per cent and there is no compression how many cubic
feet of air per hour per horsepower will be needed? What, with complete compression?
Prob. 8. A manufacturer rates his 44x42in. doubleacting engine with a speed
of 100 R.P.M. at 1000 H.P. when running noncondensing, initial pressure 70 Ibs.
per square inch gage and cutoff i. No clearance is mentioned and nothing said about
manner of expansion. Assuming s = 1.0646, find on what clearance basis this rating is
made.
Prob. 9. The water supply of a town is supplied by a directacting noncondensing
pump with two cylinders, each 24x42 ins., with 10 per cent clearance, and no com
pression, initial pressure being 100 Ibs. per square inch gage. What must be the size of the
fcteam cylinder of a crankandflywheel pump with 6 per cent clearance to give the same
jhorsepower on the same steam and exhaust pressures with a cutoff of ? Speed in
.each case to be 50 R.P.M.
NOTE: 8 for 100 Ibs. gage =.262, 8 for 15 Ibs. abs.=.0387.
7. Action of Fluid in Multipleexpansion Cylinders. General Description
Structure and Processes. When steam, compressed air, or any other high
jressure working fluid is caused to pass through more than one cylinder in
jries, so that the exhaust from the one is the supply for the next, the engine
5, in general, a multipleexpansion engine, or more specifically, a compound
n the operations are in two expansion stages, triple for three, and quadruple
For four stages. It must be understood that while a compound engine is one
which the whole pressurevolume change from initial to back pressure takes
>lace in two stages, it does not necessarily follow that the machine is a two
jylinder one, for the second stage of expansion may take place in two cylinders,
each of which, half of the steam is put through identical operations; this
226 ENGINEERING THERMODYNAMICS
would make a threecylinder compound. Similarly, tripleexpansion engines,
while they cannot have less than three may have four or five or six cylinders
Multiple expansions engine, most of which are compound, are of two classes
with respect to the treatment and pressurevolume changes of the steam, first
wthout receiver, and second, with receiver. A receiver is primarily a chamber
large in proportion to cylinder volumes, placed between the high and lowpres
sure cylinders of compounds or between any pair of cylinders in triple or
quadruple engines, and its purpose is to provide a reservoir of fluid so that
the exhaust from the higher into it, or the admission to the lower from it, will
be accomplished without a material change of pressure, and this will be accom
plished as its volume is large in proportion to the charge of steam received
by it or delivered from it. With a receiver of infinite size the exhaust line of
a highpressure cylinder discharging into it will be a constantpressure line,
as will also the admission line of the lowpressure cylinder. When, however,
the receiver is of finite size highpressure exhaust is equivalent to increasing
the quantity of fluid in the receiver of fixed volume and must be accompanied
by a rise of pressure except when a lowpressure cylinder may happen to be
taking out fluid at the same rate and at the same time, which in practice never
happens. As the receiver becomes smaller in proportion to the cylinders, the
pressure in it will rise and fall more for each highpressure exhaust and low
pressure admission with, of course, a constant average value. The greatest
possible change of pressure during highpressure exhaust and lowpressure admis
sion would occur when the receiver is of zero size, that is when there is none at all,
in which case, of course, the highand lowpressure pistons must have synchronous
movement, both starting and stopping at the same time, but moving either in
the same or opposite directions. When the pistons of the noreceiver compound
engines move in the same direction at the same time, one end of the high
pressure cylinder must exhaust into the opposite end of the low; but with
oppositely moving pistons, the exhaust from high will enter the same end of the
low. It is plain that a real receiver bf zero volume is impossible, as the connect
ing ports must have some volume and likewise that an infinite receiver is equally
impracticable, so that any multipleexpansion real engine will have receivers
of finite volume with corresponding pressure changes during the period when a
receiver is in communication with a cylinder. The amount of these pressure
changes will depend partly on the size of the receiver with respect to the cylin
ders, but also as well, on the relation between the periods of flow into receiver,
by highpressure exhaust and out of it, by lowpressure admission, which
latter factor will be fixed largely by crank angles, and partly by the settings
of the two valves, relations which are often extremely complicated.
For the purpose of analysis it is desirable to treat the two limiting cases
of no receiver and infinite receiver, because they yield formulas simple enough to
be useful, while an exact simple solution of the general case is impossible. These
simple expressions for hypothetical cases which are very valuable for estimates
and approximations are generally close to truth for an actual engine especially
if intelligently selected and used.
WORK OF PISTON ENGINES 227
Receivers of steam engines may be simple tanks or temporary storage
hambers or be fitted with coils or tubes to which live or highpressure steam
s supplied and which may heat up the lower pressure, partly expanded steam
mssing from cylinder to cylinder through the receiver. Such receivers are
cheating receivers, 'and as noted, may heat up the engine steam or may evapo
ate any moisture it might contain. As a matter of fact there can be no
leating of the steahi before all moisture is first evaporated, from which
t appears that the action of such reheating receivers may be, and is quite
iomplicated thermally, and a study of these conditions must be postponed
ill a thermal method of analysis is established. This will introduce no
serious difficulty, as such reheating receivers assist the thermal economy
)f the whole system but little and have little effect on engine power, likewise
ire now little used. Reheating of air or other gases, as well as preheating
'hem before admission to the highpressure cylinder is a necessary practice,
when the supply pressure is high, 'to prevent freezing of moisture by the gases,
which get very cold in expansion if it be carried far. This is likewise, however,
a thermal problem, not to be taken up till later.
Multipleexpansion engines are built for greater economy than is possible
n simple engines and the reasons are divisible into two classes, first mechanical,
and second thermal. It has already been shown that by expansion, work is
obtained in greater amounts as the expansion is greater, provided, of course,
xpansion below the back pressure is avoided, and as high initial and low
Dack pressures permit essentially of most expansion, engines must be built
capable of utilizing all that the steam or compressed gas may yield. If
ateam followed the logarithmic law of expansion, pressure falling inversely
with volume increase, then steam of 150 Ibs. per square inch absolute
expanding to 1 Ib. per square inch absolute would require enough ultimate
cylinder space to allow whatever volume of steam was admitted up to cutoff
to increase 150 times. This would involve a valve gear and cylinder structure
capable of admitting T J<j = .0067 of the cylinder volume. It is practically
impossible to construct a valve that will accurately open and close in this
necessarily short equivalent portion of the stroke. This, however, is not the
worst handicap even mechanically, because actual cylinders cannot be made
without some clearance, usually more than 2 per cent of the displacement and
n order that any steam might be admitted at all, the clearance in the example
would have to be less than .67 per cent of the total volume, which is quite
impossible. These two mechanical or structural limitations, that of admission
valve gear and that of clearance limits, supply the first argument for multiple
expansion engines, the structure of which is capable of utilizing any amount
of expansion that high boiler pressure and good condenser vacuum make
available. For, if neglecting clearance, the lowpressure cylinder had ten
iimes the volume of the high, then the full stroke admission of steam to the high
followed by expansion in the low would give ten expansions, while admission
k) the high for ^ of its stroke would give 15 expansions in it, after which this
inal volume would increase in the low ten times, that is, to 150 times the original
22S ENGINEERING THERMODYNAMICS
volume, and cylinder admission of ^ or 6.7 per cent is possible with ordinary
valve gears, as is also an initial volume of 6.7 per cent of a total cylinder
volume, even with clearance which in reasonably large engines may be noi
over 2 per cent of the whole cylinder volume.
It is evident that the higher the initial and the lower 'the back pressure*
the greater the expansion ratio will be for complete expansion, and as in steair
practice boiler pressures of 225 Ibs. per square inch gage or approximately 24(
Ibs. per square inch absolute with vacuum back pressures as low as one or ever
half a pound per square inch are in use, it should be possible whether desirable
or not, to expand to a final volume from 250 to 500 times the initial in round num
bers. This is, of course, quite impossible in simple engine cylinders, and as i1
easy with multiple expansion there is supplied another mechanical argument foi
staging. Sufficient expansion for practical purposes in locomotives and lane
engines under the usually variable load of industrial service is available for ever
these high pressures by compounding, but when the loads are about constant
as in waterworks pumping engines, and marine engines for ship propulsion
triple expansion is used for pressures in excess of about 180 Ibs. gage.
Use of very high initial and very low back pressures will result in simple
engines, in a possibility of great unbalanced forces on a piston, its rods, pirn
and crank, when acting on opposite sides, and a considerable fluctuation in tan
gential turning force at the crank pin. Compounding will always reduce the
unbalanced force on a piston, and when carried out in cylinders each of whict
has a separate crank, permits of a very considerable improvement of turning
effort. So that, not only does multiple expansion make it possible to utilize
to the fullest extent the whole range of high initial and low back pressures
but it may result in a better force distribution in the engine, avoiding shocks
making unnecessary, excessively strong pistons, and rods and equalizing turn
ing effort so that the maximum and minimum tangential force do not deparl
too much from the mean.
The second or thermal reason for bothering with multipleexpansion com
plications in the interest of steam economy is concerned with the preventior
of steam loss by condensation and leakage. It does not need any elaborate
analysis to show that lowpressure steam will be cooler than highpressure
steam and that expanding steam in a cylinder has a tendency to cool the
cylinder and piston walls, certainly the inner skin at least, so that aftei
expansion and exhaust they will be cooler than after admission; but as
admission follows exhaust hot live steam will come into contact with coo'
walls and some will necessarily condense, the amount being smaller the less
the original expansion; hence in any one cylinder of a multipleexpansior
engine the condensation may be less than a simple engine with the same range
of steam pressures and temperatures. Whether all the steam condensatior
during admission added together will equal that of the simple engine or no1
i^ another .question. There is no doubt, however, that as the multiple expan
sion engine can expand usefully to greater degree than a simple engine, anc
so cause a lower temperature by expansion, that it has a greater chance tc
WORK OF PISTON ENGINES 229
eevaporate some of the water of initial condensation and so get some work
ut of the extra steam so evaporated, which in the simple engine might have
emained as water, incapable of working until exhaust opened and lowered
he pressure, when, of course, it could do no good. It is also clear that steam
>r compressedair leakage in a simple engine is a direct loss, whereas in a
:ompound highpressure cylinder leakage has at least a chance to do some
vork in the lowpressure cylinder. The exact analysis of the thermal reasons
or greater economy is complicated and is largely concerned with a study of
;team condensation and reevaporation, but the fact is, that multipleexpan
sion engines are capable of greater economy than simple. The thermal analysis
nust also consider the influence of the reheating receiver, the steamjacketed
vorking cylinder, and the use of superheated steam, their effects on the pos
>ible work per pound of steam and the corresponding quantity of heat expended
;o secure it, and for air and compressed gas the parallel treatment of pre
leating and reheating.
To illustrate the action of steam in multipleexpansion engines some indi
3ator cards are given for a few typical cases in Figs. 63 to 66, together
with the combined diagrams of pressurevolume changes of the fluid in all
3ylinders to the same scale of pressures and volumes, which, of course, makes
ihe diagram look quite different, as indicator cards are usually taken to the
same base length, fixed by the reducing motion, and to different pressure
Scales, to get as large a height of diagram as the . paper will permit. Fig. 63
bows four sets of cards taken from an engine of the compound noreceiver
jype, namely, a Vauclain compound locomotive. In this machine there are
f\vo cylinders, one high pressure and one low, on each side, the steam from
he high pressure exhausting directly into the lowpressure cylinder so that
;the only receiver space is made up of the clearance and connecting passages
I between the cylinders. Starting with set A, the cards show a decreasing
igh pressure cutoff of 76 per cent in the case of set A to 54 per cent in the
iase of set D. The letters A, B, C and D refer in each case to admission, cut
, release and compression, the use of primed letters denoting the lowpressure
cylinder.
In set A the highpressure admission line AFB may be considered as made
lp of two parts, the part AF representing pressure rise at constant volume,
wh ; ch is the admission of steam to the clearance space at dead center to raise the
"essure from that at the end of compression to that of boiler pressure. From
m to B admission occurred at constant pressure, steam filling the cylinder
lolume as the piston moved outward. At B cutoff or closure of the steam
jalve occurred and the steam in the cylinder expanded. At C, release or open
4f the exhaust valve of the highpressure cylinder occurred and the admission
Valve of the lowpressure cylinder opened, the steam dropping in pressure until
the pressure in both high and lowpressure clearance became equal, and then
mpanding in both cylinders, as the exhaust from the high and admission to
j the low occurred, the exhaust line CD of the high pressure and the admission
:line F'B' of the low pressure being identical except for the slight pressure drop
230
ENGINEERING THERMODYNAMICS
(A)
(B)
(D)
FIG. 63. Set of Indicator Cards from Vauclain Locomotive Illustrating the Noreceiver
Compound Steam Engine.
WOKK OF PISTON ENGINES 231
in the passages between the high and the lowpressure cylinders. At D the
highpressure exhaust valve closed and compression of the steam trapped in
the highpressure cylinder occurred to point A, thus closing the cycle. From
point D' in the lowpressure cylinder, which corresponds to D in the high
pressure, no more steam was admitted to the lowpressure cylinder. What
steam there was in the low expanded to the point C" when the exhaust valve
opened and the pressure dropped to the back pressure and the steam was
exhausted at nearly constant back pressure to D', when the exhaust valve closed
and the steam trapped in the cylinder was compressed to A', at which point
steam was again admitted and the cycle repeated.
In set B the cycle of operation is exactly the same as in set A. In set C
the cycle is the same as in A, but there are one or two points to be especially
noted, as they are not present in set A. The admission line of the highpres
sure cylinder is not a constant pressure, but rather a falling pressure one, due to
throttling of the steam, or "wire drawing," as it is called, through the throttle
valve or steam ports, due to the higher speed at which this card was taken.
It will also be noticed that the compression pressure is higher in this case, due
to earlier closing of the exhaust valve, which becomes necessary with the type of
valve gear used, as the cutoff is made earlier. In the lowpressure card it will
be seen that the compression pressure is greater than the admission pressure
and hence there is a pressure drop instead of rise on admission. In set D the
peculiarities of C are still more apparent, the compression in highpressure
cylinder being equal to admission pressure and above it in the lowpressure
cylinder. The wire drawing is also more marked, as the speed was still
higher when this set of cards was taken.
In Fig. 64, one set of the cards of Fig. 63 is redrawn on crosssection
paper and then combined. Cards taken from the different cylinders of a
multipleexpansion engine will in nearly all cases have the same length, the great
est that can be conveniently handled by the indicator, and will be to two different
pressures scales, in as much as that indicator spring will be chosen for each
cylinder which will give the greatest height of card consistent with safety to
 the instrument. To properly compare the cards they must be reduced to the
I same pressure scale, and also to the same volume scale. As the lengths
I represent volumes, the ratio of the two volume scales will be as that of the
j cylinder volumes or diameters squared. Hence, the length of the highpressure
? card must be decreased in this ratio or the low increased. As a rule it is found
t more convenient to employ the former method. When the cards have been
reduced to a proper scale of pressures and volumes the clearance must be
> added to each in order that the true volume of the fluid may be shown.
I The cards may now be placed with these atmospheric lines and zero volume
I lines coinciding and will then appear in their true relation. In this case the
cylinder ratio was 1.65, the indicator springs 100 Ibs. and 70 Ibs. respectively
and clearance 5 per cent in each cylinder.
The steps in combining the cards were as follows: The zero volume lines were
first drawn perpendicular to the atmospheric line and at a distance from the end of
232
ENGINEERING THERMODYNAMICS
the card equal to the length of the card times the clearance. PV axes were laid
off and a line drawn parallel to the zeropressure line at a distance above it equal to
14.7 Ibs. to scale of combined diagram. This scale was taken to be that of the
highpressure diagram. A number of points A'B'C', etc., were then chosen
on the lowpressure card, and the corresponding points a' b'c', etc., plotted by
making the distances of a', &', etc., from the zerovolume line equal to thoce
of A',B', etc., and the distances of the new points above the atmosphere .7
the distances of the original. By joining the points as plotted, the new diagram
for the lowpressure card was formed. The highpressure card was then redrawn
c
A'
FIG. 64. Diagram to Show Method of Combining the High and Lowpressure Cylinder
Indicator Cards of the Noreceiver Compound Engine.
by taking a number of points A, B, C, etc., and plotting new points a, b, c,
etc., so that the distances of a, 6, c, etc., from the zerovolume line were the
l.bo
distances of A, B, C, etc., while the distances of new points above the atmos
pheric line were the same as for the original points.
In Fig. 65 are shown two cards from a compound steam engine with receiver.
Diagram A shows the cards as taken, but transferred to crosssection paper for
ease in combining, and with the zerovolume axis added. On the highpressure
card admission occurred practically at constant volume, piston being at rest
at dead center, at A, bringing the pressures in the cylinder up to the initial
pressure at B. Admission continued from B to C at nearly constant pressure,
the piston moving slowly with correspondingly small demand for steam and
consequently little wire drawing. From C to D the piston is moving more
rapidly and there is in consequence more wire drawing, admission being no
longer at constant pressure. At D the steam valve closes and expansion occurs,
to E y whefe release occurs, the pressure falling to that in the receiver. From
F to G exhaust occurs with increase of pressure due to the steam being forced
into the receiver, (receiver} decreasing H.P. cyl.vol.) while from G to H the
WORK OF PISTON ENGINES
233
pressure falls, due to the lowpressure cylinder taking steam from the receiver
and consequently volume of receiver, (receiver + increasing L.P. cyl.vol.+
decreasing H.P. cyl.vol.) increasing. At H exhaust closes, a very slight com
pression occurring from H to A .
On the lowpressure card, admission occurred at A' and continued to B' at
constant volume, the piston being on dead center as from A to B in highpressure
cylinder. From B' to C" admission occurred with falling pressure due to increase
in receiver volume, (receiver f increasing L.P. cyl.vol.), and from C' to D'
admission still took place, but with less rapidly falling pressure, as highpres
sure cylinder is now exhausting and receiver volume, (receiver + increasing
L.P. cyl. vol. + decreasing H.P. cyl.vol.) was receiving some steam as well as
! FIG. 65. Indicator Cards from a Compound Engine with a Receiver, as Taken and as
Combined.
Delivering. At D' admission ceased and expansion took place to E' where
felease occurred, the pressure falling to the back pressure and continuing from
F' to G f , where the exhaust valve closed and compression took place to A',
thus completing the cycle. At H f leakage past the exhaust valve was so great
as to cause the curve to fall off considerably from H' to A', instead of con
mnuing to be a true compression curve, ending at /, as it should have done.
iffhe combined diagrams are shown in B.
In Fig. 66 are shown a set of three cards from a tripleexpansion pumping
ingine with large receivers and cranks at 120. In diagram A the cards are
Hiown with the same length and with different pressure scales as taken, but
with the zero volume line added and transferred to crosssection paper. On
the highpressure card admission occurred at A, causing a constant volume
ressure rise to B, the piston being at rest with the crank at dead center. From
P to C admission occurred at nearly constant pressure to C, where steam was
234
ENGINEERING THERMODYNAMICS
cut off and expansion took place to D. At this point release occurred, the
pressure dropping at constant volume to E with the piston at rest. From E
exhaust took place with slightly increasing pressure, since the intermediate
cylinder was taking no steam, the intermediate piston being beyond the point
of cutoff. The pressure rise is slight, however, due to the size of the receiver,
which is large compared to the cylinder. At twothirds of the exhaust stroke,
point F, the back pressure became constant and then decreased, for at this
point the speed of the intermediate piston increased and the receiver pressure
fell. At G exhaust closed and a slight pressure rise occurred to A, due to the
restricted passage of the closing exhaust valve. On the intermediate card
At H
B"
Atn
Atm
FIG. 66. Indicator Cards from a Tripleexpansion Engine with Receiver as Taken and
Combined.
admission occurred at A', the pressure rising to B'. From B f the admissio
was at nearly constant pressure to X while the piston speed was low and the
at a falling pressure to C". Pressure was falling, since the steam was supplie
from a finite receiver into which no steam was flowing during intermediati
admission. At C" cutoff occurred and steam expanded to D', where relea:
took place, and the steam was exhausted. As in the case of the highpressu j
cylinder the back pressure was rising for twothirds of the stroke, since tl<
steam was being compressed into the receiver or rather into a volume ma<(
up of receiver and intermediate cylinder volume, which is, of course, a decreoj
ing one, since the cylinder volume is decreasing. At twothirds of the stro h
the lowpressure cylinder begins to take steam and the receiver volume I
i
WOEK OF PISTON ENGINES 235
now increased, inasmuch as it was made up of the receiver portion of the inter
mediate cylinder and a portion of the lowpressure cylinder, and the low
pressure cylinder volume increased faster than intermediate decreased for the
same amount of piston travel. At G' exhaust closed and a slight compression
occurred to A', thus completing the cycle.
On the lowpressure card admission occurred at A" and the pressure rose at
constant volume to B", and then admission continued first at constant pressure
and then falling, as in the intermediate cylinder, to the point of cutoff
at C" '. From here expansion took place to D n '. At this point the exhaust
valve opened, the pressure fell nearly to back pressure at E", and the steam
was exhausted at practically constant back pressure to G" ', where the exhaust
valve closed and there was compression to A", thus completing the cycle.
The combined diagram is shown in B.
Prob. 1. In Fig. 67 are shown six sets. of indicator cards from compound engines.
The cylinder sizes and clearances are given below. Explain the cylinder events and the
shape of lines for each card and form a combined diagram for each set.
No. 1. From a four valve Corliss engine, 26x48 ins., with 3 per cent clearance in
each cylinder.
No. 2. From a single valve engine, 12x20x12 ins., with 33 per cent clearance in
highpressure cylinder and 9 per cent in low.
No. 3. From a fourvalve Corliss engine 22x44x60 ins., with 2 per cent clearance
in the highpressure cylinder and 6 per cent in low.
No. 4. From a single valve engine 18 X30X16 ins., with 30 per cent clearance in the
highpressure cylinder and 8 per cent in the low.
No. 5. From a single valve engine 11^X18^X13 ins., with 7 per cent clearance in
the high and 10 per cent in the low.
No. 6. From a double valve engine 14 X 28 X2^ ins.,, with 3.5 per cent clearance in
the highpressure cylinder and 6.5 per cent in the low.
Prob. 2. In Fig. 68 are shown four sets of indicator cards from tripleexpansion
marine engines. The cylinder sizes and clearances are given below. Explain the cylinder
events and the shape of the lines of each card and form a combined diagram of each set.
No. 1. From the engine of a steamship, cylinders 21.9x34x57 ins. X39 ins. with
6 per cent clearance in each and fitted with simple slide valves.
No, 2. From an engine 20x30x50x48 ins.
No. 3. From the engine of a steamship with cylinders 22x35x58x42 ins.,
assume clearance 7 per cent in each cylinder.
No. 4. From the engine of the steamer "Aberdeen, " with cylinders 30 X45 X70 X54
ins., and with 4 per cent clearance in the high, 7 per cent in the intermediate, and 8 per
jj&ent in the low.
Prob. 3. In Fig. 69 are shown some combined cards from compound engines.
Explain the cylinder events and the shape of the lines and reproduce the indicator cards.
Prob. 4. In Fig. 70 are ahown some combined cards from tripleexpansion engines.
'Draw the individual cards and explain the cylinder events and shape of lines!
8. Standard Reference Cycles or PV Diagrams for the Work of Expansive
Fluids in Twocylinder Compound Engines. The possible combinations of
jadmission with all degrees of expansion for forward strokes and of exhaust with
236 ENGINEERING THERMODYNAMICS
all degrees of compression for back strokes, with and without clearance, in
each of the two cylinders of the compound engine, that may have any volume
relation one to the other and any size of receiver between., and finally, any sort
of periodicity of receiver receipt and discharge of fluid, all make possible a
100
GO
20
0
No. 2
lOi
o
10
No. 3
No. 4
40
20
0
10
40
20
0
JO
100
60
20
0
120
0
No. 6
10
0
FIG. 67. Six Sets of Compound Engine Indicator Cards.
very large number of cycles. In order that analysis of these conditions of work
ing may be kept within reasonable space, it is necessary to proceed as was
done with compressors and simple engines, concentrating attention on sucl
type forms as yield readily to analytical treatment and for which the formula;
WORK OF PISTON ENGINES
are simple even if only approximate with respect to actual engines, but, of
course, keeping in mind the possible value of the formulas, as those that teach
FIG. 68. Four Sets of Tripleexpansion Engine Indicator Cards.
no principles or fail to assist in solving problems must be discarded as useless.
The work that fluids under pressure can do by losing that pressure is no
238
ENGINEERING THERMODYNAMICS
different in compound than in simple engines, if the fluid has a chance to do
what ii can. Provided the structure is such as will not interfere with the com
pleteness of the expansion, and no fluid is wasted in filling dead spaces without
1301
200
100
100
No. 7
No. 8
Fio. 69. Combined Diagrams of Compound Engines.
working, then the work per cubic foot or per pound of fluid is the same for simple
compound and triple engines. Furthermore, there is a horsepower equiv
alence between the simple and rorn pound, if, in the latter case the steanr
\\oicu OK PISTON I:\<JI\KS
Up t. ruf
no
,.
100
r
onlv in flu I mr
'\,l :>: l( <i \ < :r :i
X.. 1
N.. ft
l'i,, ,0 ( 'uinhln..l I,,,,MM.M.. ..i I,,,, I, , tuiiiMnii I'',..,.,....'
il ..II IIH'MII ... in.lrt TliiM flu.ill.l lir rl.'MI 1 1 <Uli H ruiii 1:1 1 \^ t 71,
MM, I II III lii ,11. i, .i. . IllillfJ lll< . II I , 11,,,. I. , ur.n, VN ill,.. ill
240
ENGINEERING THERMODYNAMICS
clearance and with complete expansion, the volume admitted, AB, expand
to the back pressure on reaching the full cylinder volume DC, and exhausl
at constant back pressure, the work represented by the area A BCD. It shoul
be clear that no difference will result in the work done if a line be drawn acros
the work area as in Fig. 7 IB, all work done above the line HG to be develope
in the highpressure cylinder and that belowjn the low. This is merely equn
ulent to saying that a volume of steam AB is admitted to the highpressui
cylinder expands completely to the pressure at G on reaching the full higl
pressure cylinder volume, after which it exhausts at constant pressure (int
a receiver of infinite capacity), this same amount being subsequently admitte
without change of pressure to the lowpressure cylinder, when it again expanc
completely. Thus, it appears that the working of steam or compressed a
Vol. Admitted to Cylinder
\
Displacement Volume
I .Ad:
Vol. Admitted to H
H.P. Displ icement
B
Volume Admitted to L.P. Cyl.
Cyl rider
L.P. Displacement 
FIG. 71. Diagram to Show Equality of Work for Expansion in Onecylinder Simple and
Twocylinder Compound Engines for the Same Rate of Expansion.
in two successive cylinders instead of one will in no way change the maximu
amount of work a cubic foot supplied can do, the compounding merely maki
it easier to get this maximum. In simple engine cases, Fig. 71A, the cutoff in p
AB
cent of stroke is 100 X 7^, which is a very small value, leaving but little time
.DC
open and close the admission valve, whereas in the compound case the per ce
cutoff in the highpressure cylinder is lOOXTTT^and in the lowpressure cylind
100X ,^, which are quite large enough ratios to be easily managed with ordina
L/L>
valve gc.i
Compounding docs, however, introduce possibilities of loss not present
the sinjilcslugc expansion, if the dimensions or adjustments are not rig
which may be classed somewhat improperly as receiver losses, and these
WORK OF PISTON ENGINES
241
of two kinds, one due to incomplete
expansion in the high and the other
to overexpansion there. Thus, in
Fig. 72, if ABCEFGDA represent a
combined compound diagram for
the case of complete expansion in
the highpressure cylinder continued
in the low without interruption but
incomplete there, DC represents the
volume in the lowpressure cylinder
at cutoff, and at the same time the
total highpressure cylinder volume.
If now, the lowpressure cutoff
be made to occur later, Fig. 73,
then the volume that the steam _
would occdpy when expansion began F 7
in the lowpressure cylinder is rep
resented by D'C'. This adjust
ment could not, of course, change the
\
\
WIG. 73. Diagram to Show Effect of Lengthening
[ L.P. Cutoff Introducing a Receiver Loss Due to
? Incomplete Highpressure Cylinder Expansion.
D 2 j V
!. Diagram to Show Correct Lowpressure
Cutoff for No Receiver Loss.
highpressure total volume DC, so
that at release in the highpressure
cylinder the pressure would drop
abruptly to such a value in the
receiver as corresponds to filling
the low pressure up to its cutoff,
and work be lost equal to area
CC'C".
A shortening of the lowpressure
cutoff will have an equally bad
effect by introducing negative work
as indicated in Fig. 74, in which
the L.P. cutoff volume is reduced
from DC to D'C' Expansion in the
high pressure proceeds as before till
the end of the stroke, at which time
it has a volume DC greater than
the low pressure can receive D'C',
hence the receiver pressure must
rise to such a value as will reduce
the volume the required amount,
introducing the negative work CC'C" .
Changes of lowpressure cutoff may,
therefore, introduce negative work,
change the receiver pressure and,
of course, modifiy the distribution
of work between high and low, but
242
ENGINEERING THERMODYNAMICS
these same effects might also have resulted from changes of highpressure
cutoff or of cylinder ratio.
For such conditions as have been assumed it seems that compounding
does not increase the work capacity of fluids, but may make it easier to realize
this capacity, introducing at the same time certain rather rigid relations between
cutoffs and cylinder volumes as necessary conditions to its attainment. It
can also be shown that the same proposition is true when there are clearance and
compression, that is, in real cylinders and when the receiver is real or not
infinite in size, or when the exhaust of high and admission of low, are not con
stantpressure lines. The former
needs no direct proof, as inspec
tion of previous diagrams makes
it clear, but the latter requires
some discussion.
A real receiver of finite size
is at times in communication
with the highpressure cylinder
during its exhaust, and at other
times with the lowpressure
cylinder during admission, and
these two events may take place
at entirely independent times, be
simultaneous as to time, or over
lap in all sorts of ways. Suppose
the periods to be independent
and there be no cylinder clear
ance, then at the beginning of
highpressure exhaust two sepa
rate volumes of fluid come
together, the contents of both
the highpressure cylinder and the
receiver, and this double volume
is compressed by the H.P. piston
into the receiver, in which case the
highpressure exhaust would take
place with rising pressure . Follow
ing this will come lowpressure admission, during which the volume of fluid in the
receiver expands into the lowpressure cylinder up to its cutoff, and if the same
volume is thus taken out of the receiver as entered it previously, lowpressure
admission will take place with falling pressure, the line representing it exactly
coinciding with that for the highpressure exhaust. Independence of H.P.
cylinder exhaust and L.P. cylinder admission, as to time, may result in a cycle
such as is represented in Fig. 75 for the case of no cylinder clearance. On this
diagram the receiver line is DC, an expansion or compression line referred to a
second axis of volumes KJ, placed away from the axis of purely cylinder volumes
I i , .71. Diagram to Show Effect of Shortening L.P.
Cutoff, Introducing a Receiver Loss Due to
Overexpansion in the Highpressure Cylinder.
WORK OF PISTON ENGINES
243
>y the distance LD, equal to the receiver volume to scale. All diagram points
; are referred to the axis A I except those on the line DC.
This same case of time independence of H.P. exhaust and L.P. admission
yields quite a different diagram when the cylinder clearance is considered.
Such a case is represented by the diagram, Fig. 76, which also serves to illustrate
the effect of incomplete expansion and compression as to equalization of
receiver with cylinder pressures. At highpressue release the volume of fluid
in the H.P. cylinder is ML and its pressure is LR. This is about to come
into communication with the receiver volume ON from which the lowpres
sure cylinder finished taking fluid and which is, therefore, at the same pressure
K',
8r
. Receiver.
V.olu
FIG. 75. Diagram to Illustrate Variable Receiver Pressure for the Case of Independent
Highpressure Exhaust and Lowpressure Admission with Zero Clearance.
as the L.P. cylinder cutoff KS. The question, therefore, is what will be
jthe pressure at P in both H.P. cylinder and receiver when LM cu.ft. of fluid
ftt LR pressure combines with ON cu.ft. at pressure KS, and together occupy
a volume ON+LM. By hypothesis the pressure after mixture is
(first volume Xits pressure) + (second volume X its pressure)
sum of volumes
From this or the graphic construction following, the point P is located. If the
highpressure expansion had continued to bring LQ to the receiver pressure
KS, it would reach it at X. At this hypothetic point there would be a volume
NX in the H.P. cylinder to add to the volume ON in the receiver at the same
pressure, resulting in OX cu.ft. This fluid would have a higher pressure at the
[lesser volume of receiver and H.P. cylinder and the value is found by a compres
244
ENGINEERING THERMODYNAMICS
line through X, XPAT referred to the receiver axis. This same line is also
the exhaust of the H.P. cylinder from P to A . A similar situation exists at admis
sion to the L.P. cylinder as to pressure equalization and location of admission
line. At the end of the L.P. compression there is in the L.P. cylinder FE_ cu.ft.
at pressure EH, to come into communication with the receiver volume CB cu.ft.
at pressure BG, that at which H.P. exhaust ended. Producing the L.P. com
pression line to /, the volume BI is found, which, added to receiver, results
in no pressure change. An expansion line, referred to the receiver axis through 7,
fixes the equalized pressure J and locates the L.P. admission line JK, which, it
must be observed, does not coincide with the H.P. exhaust.
So far only complete independence of the time of H.P. exhaust and L.P.
admission have been considered, and it is now desirable to consider the diagram
FIG. 76. Diagram to Illustrate Variable Receiver Pressure for the Case of Independent High
pressure Exhaust and Lowpressure Admission with Clearance.
matic representation of the results of complete coincidence. Such cases occur
in practice with the ordinary tandem compound stationary steam engine and
twincylinder singlecrosshead Vauclain compound steam locomotive. In the
latter structure both pistons move together, a single valve controlling both
cylinders, exhaust from high taking place directly into low, and for exactly
equal coincident time periods. The diameter of the lowpressure cylinder being
greater than the high, the steam at the moment of release suffers a drop in
pressure in filling the lowpressure clearance, unless, as rarely happens, the pres
sure in the low is raised by compression to be just equal to that at H.P. release.
After pressure equalization takes place, highpressure exhaust and lowpres
sure admission events are really together a continuation of expansion, the
volume occupied by the steam at any time being equal to the difference between
WORK OF PISTON ENGINES
245
the two cylinder displacements and clearances up to that point of the stroke.
Before this period of communication, that is, between highpressure cutoff and
release, the volume of the expanding fluid is that of the highpressure displace
ment up to that point of the stroke, together with the highpressure clearance.
After the period of communication the volume of the expanding fluid is that
of the low pressure cylinder up to that point of the stroke, together with the low
pressure clearance, plus the highpressure displacement not yet swept out,
and the highpressure clearance.
These fluid processes cannot be clearly indicated by a single diagram,
because a diagram drawn to indicate volumes of fluid will not show the volumes
in the cylinders without distortion. If there be no clearance, Fig. 77 will
assist in showing the way in which two forms of diagram for this purpose
are derived. Referring to Fig. 77A, the volume A B admitted to the high
pressure cylinder expands in it until it occupies the whole H.P. cylinder volume
Z)C. At this point expansion proceeds in low and high together, with decreasing
volumes in high and increasing in low until the lowpressure cylinder volume is
attained at E. The line BCE then indicates the pressures and volumes of the
fluid expanding, but does not clearly show the volume in either cylinder between
C and E, with the corresponding pressure. It is certain, however, that when
the volume in the H.P. cylinder becomes zero the pressure must have fallen
to a value the same as that in the low when the fluid completely fills it,
or P,=P..
As the highpressure piston returns, on the exhaust stroke, the low
iressure piston advances an equal distance, on its admission stroke, sweep
ig through a greater volume than the high pressure, in the ratio of lowpres
re to highpressure displacements. If at any point of the stroke the volume
taining in the highpressure cylinder be x, and the high and lowpressure
displacements be respectively DI, and I>2, then (Z>i x) is the volume swept
out by highpressure piston, x the volume remaining in it', and (Dix)
the volume swept in by the lowpressure piston. Then the total volume
still in the two cylinders is, for a point between C and E,
ince the equation of the curve CE is, PV = P e Ve = PcDi, the value of V may
Tie substituted, giving P \D 2 x(j^l} =P c I>i,= constant. Dividing by
I L \ i / J
^1 )this becomes PI T^^ x\=P\ ^^ z =a new constant, so
246 ENGINEERING THERMODYNAMICS
that if a new axis LM be laid off on B, 07= from the axis GP
point on FC will be distant from the new axis LM an amount I (n 2 !_j^ ~ ) x \
as the product of this distance and the pressure P, is constant, the curve FC
is an equilateral hyperbola referred to the axis LM. Therefore Fig. 77B is
G 1 2 3 M
COMBINED DIAGRAM TO ONE SCALE
FIG. 77. Diagram to Illustrate the Compound Engine Cycle with Noreceiver, and Exact
Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance.
the pressurevolume representation of the entire cycle of the highpressure
cylinder.
In Fig. 77C is shown the corresponding pressurevolume diagram for the low
pressure cylinder, for which it may similarly be shown that the curve DE may be
WOEK OF PISTON ENGINES 24^
plotted to an axis NO at a distance to the left of the axis GP equal to the sam<
quantity,
AJZV
(295;
These diagrams, 77 A, B and C may be superposed, as in Fig. 77 E, giving
one form of combined diagram used for this purpose, and the one most nearly
comparable with those already discussed. In this diagram, the area ABCFA
represents the work of the highpressure cylinder, and DEIHD, the work of
the lowpressure cylinder. Together, they equal the work of the enclosing
figure ABEIHA, and hence the work of the lowpressure cylinder must also be
represented by the area FCEIHF.
It is not difficult to show that if a vertical, CJ, be dropped from the point
C to the exhaust line HJ, the figure CFHJ, in Fig. 77D is similar to DEIH,
in Fig. 77E, reversed, but drawn to a different horizontal scale. Here the
length of the lowpressure diagram is made equal to the length of the high
: pressure diagram. The two scales of volumes are shown above and below the
figure. While this appears to be a very convenient diagram, it will be found
to be less so when clearance and compression are considered.
It may be noted that since it has been shown that the curves CF and DE
are of the same mathematical form (hyperbolic) as the expansion line CE,
they may be plotted in the same way after having in any way found the^axis.
The location of this axis may be computed as given above, or may be found
graphically by the method given in connections with the subject of clearance,
Chapter I, and shown in Fig. 77J5. Knowing two points that lie on the curve,
C and F, the rectangle CDFK is completed. Its diagonal, DK, extended, cuts
the horizontal axis GV in the point M, which is the base of the desired axis ML.
If now the axis NO and the figure DEIH, part of which is referred to this
axis, be reversed about the axis GP, Fig. 77(7, NO will coincide with ML,
Fig. 77D, and Fig. 78 results. Note that the axis here may be found graphically,
in a very simple way. Draw the vertical CK from C to the axis GV and
the horizontal DJ to the vertical IE extended. DC is then the highpressure
displacement and DJ the lowpressure displacement. Draw the two diagonals
DK and JG, extending them to their intersection X. By similar triangles it
may be shown that a horizontal line, UW, will have an intercept between these
two lines, JG arid DK } equal to the volume of fluid present in the two cylinders
combined. The intersection X is the point at which this volume would
become zero if the mechanical process could be continued unchanged to that
point, and is, therefore, on the desired axis ML extended. T being the inter
section of WU with the axis GP, when the volume UT is present in the high
pressure cylinder, TW gives the volume in the lowpressure cylinder.
When clearance and compression are considered, this diagram is changed in
many respects, and is shown in Fig. 79. The axes OP, 0V and 0V are laid out,
with OZ equal to the clearance and ZK, the displacement, of the highpressure
ENGINEERING THERMODYNAMICS
248
cylinder, and OQ and QY, clearance and displacement of the lowpressure cylinder.
It is necessary to know highpressure cutoff, =; highpressure compression,
; and lowpressure compression, =, hi addition to the initial and back
ZK. *(
pressures, hi order to lay out the diagram. The drop in pressure CD at release
is due to the coming together of (volume V c at pressure P c ), with (volume V,
at pressure P 3 ). If the volume V, (measured from axis OP) were decreased
sufficiently to raise the pressure in the lowpressure clearance to the pressure
3
A
B
L
[
\
pi
\
J
D
\
C
"^
^
^W
^
,S
T\
U
/
:
E
, *
__

*s.^
_l
>^

^.
^
^^^
^^
F
^"
\
\
^
\
\
1
^^^
\
v>
R
""* .
^^
G
\
V
K
\
1 V
j
:
<
"x.^
\
!
\
:
*.
^
\
\
^
4
FIG. 78. Diagram to Show Volume of Steam in the Cylinders of the Noreceiver Compound
Engine at any Point of the Stroke for the Case of Noclearance and Coincidence of
H.P. Exhaust with L.P. Admission.
at C, the volume would become V s , as indicated at th,e point S } and the volumes
now combined in the hypothetical condition would occupy the volume SC.
Increasing this volume to D'D, that actually occupied after communication,
the pressure would fall along the curve SD', which is constructed on KV and KT
as axes. The pressure of D is then laid out equal to the pressure at D'. To
find the axis, ML, for the curves DE and D'E', from any convenient point N
on ZA, draw the line NK extended to X. Extend HG to R, at a height equal
to that of N, and draw RQX, and through the intersection draw the desired axis
XML. The fraction of stroke completed at E' in the low pressure at cutoff must
be equal to that completed at E in the high pressure at compression, and may be
laid out graphically by projecting up from E to the point U on the line NK and
horizontally from U to W on the line RQ. Projecting down from W to the
curve at E' locates the point of effective cutoff in the lowpressure cylinder.
WORK OF PISTON ENGINES
249
kfter the supply from the highpressure cylinder has been cut off at E', the
xpansion is that of the volume in the lowpressure cylinder and its clearance, and
ence the curve E'G is constructed on OP as an axis.
While in this last case a zero receiver volume has been assumed, there is
.othing to prevent a receiver volume being interposed between H.P. and L.P.
o that common expansion takes place with a volume greater than assumed
>y so much as this volume, the effect of which is to decrease the slope of DE
nd D'E 1 '. Such receivers usually consist of the spaces included in a valve
ody and connecting passages and may be treated generally as increased L.P.
learances.
The most common of all relations between H.P. exhaust and L.P. admis
ion is, of course, that of partial coincidence of periods, as it is thus with all cross
IG. 79. Diagram to Show Volume of Steam in the Cylinders of the Noreceiver Compound
Engine at any Point of the Stroke for the Case of Clearance and Coincidence of H.P.
Exhaust with L.P. Admission.
impound and tripleexpansion engines having separate cranks for the
^dividual cylinders. For these there is no simple fixed relation between the
eriods, for, while crank angles are generally fixed in some comparatively simple
Ration, such as 90, 180 and 270 for compounds and 120 for triples, they
ie sometimes set at all sorts of odd angles for better balance or for better
irning effort. Even if the angles were known the receiver line would have
Ibe calculated point by point. When the H.P. cylinder begins to discharge into
[receiver for, say, a cross compound with cranks at 90, steam is compressed
ito the receiver, and so far the action is the same as already considered for
dependence of periods, but at near midstroke the lowpressure admission
fens while highpressure exhaust continues. This will cause the receiver
tessure to stop rising and probably to fall until the low pressure cuts off, which
250
ENGINEERING THERMODYNAMICS
may occur before the H.P. exhaust into the receiver ceases. If it does, the receive:
pressure will again rise. Exact determination of such complex receiver lines is
not often wanted, and when needed is best obtained graphically point by point
Its value lies principally in fixing exactly the work distribution between cylin
ders, which is not of great importance except for engines that are to work a1
constant load nearly all the time, such as is the case with city water work*
pumping, and marine engines. While equations could be derived for these
cases, they are not worth the trouble of derivation, because they are too cumber
some, and graphic methods are to be substituted or an approximation to be made
Four kinds of approximation are available, as follows, all of which ignon
partial coincidence of periods:
1. Receiver pressure constant at some mean value and clearance ignore
2. Receiver pressure constant at some constant value and clearance co
sidered with compression zero or complete.
3. Receiver pressure fluctuates between fixed limits as determined by
assumed size, clearance ignored.
4. Receiver pressure fluctuates between fixed limits as determined by
assumed size, clearance considered, with compression zero or complete.
These are not all of equal difficulty in solution, and the one to be used
that nearest the truth as to representation of conditions, which is usually t
most difficult, provided time permits or the information is worth the troub
Quickest work is accomplished with assumption (1) and as this is most oft
used in practical work it indicates that its results are near enough for mo
purposes.
This discussion leads, therefore, to the analytical study of the following cycl<
INFINITE RECEIVER, ZERO CYLINDER CLEARANCE.
CYCLES V, AND VI (Fig. 80).
(a) Admission at constant supply pressure to H.P. cylinder.
(6) Expansion in H.P. cyl. (may be zero) by law PV=c for (V
PF*=cfor(VI).
(c) Equalization of H.P. cyl. pressure with receiver pressu
at constant volume (may be zero).
(d) Exhaust into infinite receiver at constant pressure fro
H.P. cylinder.
(e) Equalization of H.P. cylinder pressure with supply pressu
at constant zero volume.
(f) Admission from receiver at constant receiver pressure
L.P. cylinder.
(g) Expansion in L.P cylinder (may be zero) by law PV = c f
(V); PF'=cfor (VI).
(h) Equalization of L.P. cylinder pressure with back pressu
at constant volume (may be zero).
(i) Exhaust at constant back pressure for L.P. cylinder.
(f) Equalization of L.P. cylinder pressure with receiver pressi
at constant zero volume.
H.P. CYLINDER
EVENTS
L.P. CYLINDER
EVENTS
WORK OF PISTON ENGINES
251
p
a I
a i
\

Cycles VII & VIII
ycle VII PV=C Cycle VIII PV=C
\
Cycles V & VI
Cycle V PV=C Cycle VI PVS=C
e
\
\
t
I
V
\
\
V
SB
V
o
\
<
Rec
ivt'i
Vol
line
*
l\
c
""^^
\
V
__k
5
^
^
/
x
^>
j
N
^
j
y
^^^,
* .
^>
1 ^.
t
h
i
k
V
V
a I
p
a \
B
Cycles \X&X
Cycle IX PV=C Cycle X
PV=C
f
1
Cycles XIII & XIV
\
\Cycle XIII PV=C Cycle XIV PV=C
d
5
1 &\
\
\
\
\
\
,/
\
<*
Rec
eivei
Vol
e
>
sSsd
c
S
N^
/
x v
r^
^
y
^**"*
<
^^.
x x
^
\
^
^~.
,
X"
***^
^* ^
!
S3
lr
~~^
~~_
/
9
i.
t
1
j
1
V
V
p
a 11
a
\
Cycles XI & XII
Cycle XI PV=C Cycle XII PV=C
Cycles XV & XVI
Cycle XV PV=C Cycle XVI PVS=C
l\
\
A
A
\
\
\
(>
C
iV
f C
d
\ s
\.
y
9
\
5
/
a
"
~^>,
"^s
1
^
^
\
*"*
i

~ .
1
(
i
s
s^^
i
h
r
FIG.
V V
go. Compound Engine Standard Reference Cycles or PV Diagrams.
252
ENGINEERING THERMODYNAMICS
RELATIONS
BETWEEN H.P.
AND L.P. CYLINDER
EVENTS
(1) H.P. exhaust and L.P. admission independent as to time,
coincident as to representation (except as to length).
(2) H.P. expansion line produced coincides as to representation
with L.P. expansion line.
(3) The length of the constant pressure receiver line up to the
H.P. expansion line produced is equal to the length of
the L.P, admission line.
H.P. CYLINDER
EVENTS
FINITE RECEIVER, ZERO CYLINDER CLEARANCE.
CYCLES VII, AND VIII, (Fig. 80).
(a) Admission at constant supply pressure to H.P. cylinder.
(6) Expansion in H.P. cylinder (may be zero) by law PV=c
for (VII); PV s =c for (VIII).
(c) Equalization of H.P. cylinder pressure with receiver pressure
at constant volume (may be zero) with a change of receiver
pressure toward that at H.P. cylinder release (may be
zero).
(d) Exhaust into finite receiver from H.P. cylinder at rising
pressure equivalent to compression of fluid in H.P. cylinder
and receiver into receiver by law PF=c for (VII) and
PV s =c for (VIII).
(e) Equalization of H.P. cylinder pressure with supply pressure
at constant zero volume.
L.P. CYLINDER
EVENTS
RELATION
BETWEEN H.P. AND
L.P. CYLINDER
EVENTS
(/) Admission from receiver to L.P. cylinder at falling pressure
equivalent to expansion of fluid in receiver into receiver
and L.P. cylinder together by law PV=c 'for (VII), PV S =c
for (VIII).
(0) Expansion in L.P. cylinder (may be zero) by law PV=c
for (VII); PV s =c for (VIII).
(h) Equalization of L.P. cylinder pressure with back pressure
at constant volume (may be zero).
(1) Exhaust at constant back pressure for L.P. cylinder.
(/) Equalization of L.P. cylinder pressure with receiver pressure &
at constant zero volume to value resulting from H.P.(y
exhaust.
(1) H.P. exhaust and L.P. admission independent as to time,!
coincident as to representation, except as to length.
(2) H.P. expansion line produced coincides as to representation!!
with L.P. expansion line.
(3) The length of the receiver pressure line up to the H.P.I
expansion line produced is equal to the length of the 
L.P. admission line.
WOEK OF PISTON ENGINES
253
No RECEIVER, ZERO CYLINDER CLEARANCE.
CYCLES IX, AND X, (Fig. 80).
H.P. CYLINDER
EVENTS
BOTH H.P. AND L.P.
SIMULTANEOUSLY
H.P. CYLINDER
EVENTS
L.P. CYLINDER
EVENTS
(a) Admission at constant supply pressure to H.P. Cylinder.
(6) Expansion in H.P. cylinder (may be zero) by law PV
for (IX); PF s =cfor(X).
(c) Transference of fluid from H.P.to L.P. cylinder with simul
taneous continuation of expansion until all fluid is in
L.P. cylinder and expanded to its full volume by law
PV=c for (IX); PV S =c for (X).
( (d) Equalization of H.P. cylinder pressure to the pressure of
supply.
(e) Equalization of L.P. cylinder pressure with back pressure
at constant volume (may be zero).
(/) Exhaust at constant back pressure for L.P. cylinder.
(g) Equalization of L.P. cylinder pressure to the pressure in
H.P. cylinder at the end of its expansion.
[.P. CYLINDER
EVENTS
INFINITE RECEIVER, WITH CYLINDER CLEARANCE.
CYCLES XI, AND XII, (Fig. 80).
(a) Admission at constant supply pressure to H.P. cylinder.
(6) Expansion in H.P. cylinder (may be zero) by law PV=c
for (XI); PF s =cfor(XII).
(c) Equalization of H.P. cylinder pressure with receiver pressure
at constant volume (may be zero) pressure.
(d) Exhaust into infinite receiver at constant pressure from H.P.
cylinder.
(e) Compression in H.P. cylinder to clearance volume (may be
zero) by law PV =c for (XI) ; PF S =c for (XII).
(/) Equalization of H.P. cylinder pressure with supply pressure
at constant clearance volume, may be zero.
(g) Admission from receiver at constantreceiver pressure to
L.P. cylinder.
(ft) Expansion in L.P. cylinder (may be zero) by law PV c
for (XI); P7 s =cfor(XII).
(i) Equalization of L.P. cylinder pressure with back pressure
.P. CYLINDER I at constant volume (may be zero).
EVENTS  (/) Exhaust at constant back pressure for L.P. cylinder.
(k) Compression in L.P. cylinder to clearance volume by law
PV = c for (XI) ; PV S =c for (XII) (may be zero).
(/) Equalization of L.P. cylinder pressure with receiver pres
sure at constant clearance volume without change of
receiver pressure (may be zero).
254
ENGINEERING THERMODYNAMICS
RELATIONS
BETWEEN H.P. AND
L.P. CYLINDER
EVENTS
(1) H.P. exhaust and L.P. admission independent as to til
coincident as to representation except as to length.
(2) L.P. expansion line does not coincide as to representation
with H.P. expansion line produced by reason of clearance
influence except in one special and unusual case.
(3) The length of the constantreceiver pressure line intercepted
between H.P. compression line and H.P. expansion line
produced is equal to the same intercept between L.P.
expansion line and L.P. compression line produced. This
is equivalent to the condition that the volume taken in by
low is equal to expelled by the high reduced to the same
pressure.
FINITE RECEIVER, WITH CYLINDER CLEARANCE.
CYCLES XIII, AND XIV, (Fig. 80).
H.P. CYLINDER
EVENTS
L.P. CYLINDER
EVENTS
(a) Admission at constant supply pressure to H.P. cylinder.
(6) Expansion in H.P. cylinder (may be zero) by law PV=c
for (XIII) ; PV S =c for (XIV).
(c) Equalization of H.P. cylinder pressure with receiver pres
sure at constant volume (may be zero) toward that at
H.P. cylinder release (may be zero).
(d) Exhaust into finite receiver from H.P. cylinder at rising
pressure equivalent to compression of fluid in H.P. cylinder
and receiver into receiver by law PV =c for (XIII) ; PV S =c
for (XIV).
(e) Compression in H.P. cylinder to clearance volume (may be
zero) by law PV=c for (XIII); PV S =c for (XIV).
(/) Equalization of H.P. cylinder pressure with supply pressure
at constant clearance volume.
(g) Admission from receiver to L.P. cylinder at falling pressure;
equivalent to expansion of fluid in receiver into receiver j
and L.P. cylinder together by P7=c for (XIII); PV S =c\
for (XIV).
(h) Expansion in L.P. cylinder (may be zero) by law PV =c\
for(V); PV =cfor(VI).
(i) Equalization of L.P. cylinder pressure with back pressure!
at constant volume (may be zero).
(f) Exhaust at constant back pressure for L.P. cylinder.
(k) Compression in L.P. cylinder to clearance volume by lawj
PV=c for (XI); PV'=c for (XII) (may be zero).
(/) Equalization of L.P. cylinder pressure with receiver pressure j
at constant clearance volume with change of receive! ;
pressure in direction of L.P. compression pressure (maj)
be zero).
WORK OF PISTON ENGINES
255
RELATIONS
ETWEEN H.P. AND
L.P. CYLINDER
EVENTS
(1) H.P. exhaust and L.P. admission independent as to time,
representation and length.
(2) L.P. expansion line does not coincide as to representation
with H.P. expansion line produced by reason of clearance
influence except in one special and unusual case.
(3) The highpressure exhaust and lowpressure admission lines
do not coincide as to representation by reason of clearance
influences.
(4) There is a relation between the lengths of the L.P. admission
and H.P, exhaust lines, but not a simple one.
No RECEIVER, WITH CYLINDER CLEARANCE.
CYCLES XV, AND XVI, (Fig. 80).
H.P.
CYLINDER
EVENTS
LT.
CYLINDER
EVENTS
(a) Admission at constantsupply pressure to H.P. cylinder.
(6) Expansion in H.P. cylinder (may be zero) according to law
PV=c for (XV); PV S =c for (XVI).
(c) Equalization of pressures in H.P. cylinder after expansion
with that in L.P. after compression at constant volume
(may be zero) .
Transference of fluid from H.P. to L.P. cylinder until all
fluid is in L.P. cylinder and expanded to its full volume
by same law as (6) .
Compression in H.P. cylinder to clearance volume (may be
zero) by law PV=c for (XV); PV S =c for (XVI).
(/) Equalization of pressure in H.P. cylinder with supply at
constantclearance volume (may be zero).
(g) Expansion in L.P. cylinder may be zero by law PV=c for
(XV); FF s =cfor(XVI).
(h) Equalization of pressure in L.P. cylinder with back pressure,
at constant volume (may be zero).
(i) Exhaust at constant pressure for L.P. cylinder,
(j) Compression in L.P. cylinder to clearance, may be zero by
law PV =c for (XV) ; PV 9 =c for (XVI).
(k) Equalization of L.P. cylinder pressure with H.P. cylinder
pressure.
Cycle XVII, Fig. 81, for the triple expansion is denned in the same way
.s the corresponding case of compounds Cycle V, with appropriate alterations
9 wording to account for a third or intermediate cylinder between high and low
pessure cylinders and an additional receiver. Thus, high pressure cylinder
xhausts into first, and intermediate cylinder into second receiver: inter
mediate cylinder receives its supply from first, and lowpressure cylinder from
cond receiver. This being the case, it is unnecessary to write out the cylin
mr events, noting their relation to the corresponding compound case.
256
ENGINEERING THERMODYNAMICS
9. Compound Engine with Infinite Receiver, Logarithmic Law. No Clear
ance, Cycle V. General Relations between Pressures, Dimensions, and Work.
It must be understood that the diagrams representing this cycle, Fig. 82,
indicating 'A) incomplete expansion and (B) overexpansion in both cylinders,
P
m
FIG. 81. Tripleexpansion Engine Standard Reference Diagram or PV Cycle for Infinit
Receiver.
may just as well stand for over, complete or incomplete expansion in all possibl
combinations in the two cylinders. Applying the principles already derivec
for calculating the work areas,
Highpressure cylinder work
fefcw*
(296
Lowpressure cylinder work
(297
WORK OF PISTON ENGINES 257
Total work
P d V d P V ff> . . (298)
pressure being in pounds per square foot, and volumes in cubic feet.
In theses expressions the receiver pressure P e = P d is unknown, but determinate
as it is a function of initial pressure and certain volumes, giving it the value,
merely satisfying the condition that the point E at which expansion begins
in the lowpressure cylinder must lie in the expansion line of the high. Sub
stituting this value there results
?+lQ fe PJV . (299)
This is a perfectly general expression for the work of the fluid expanding to
any degree in two cylinders in succession when the clearance is zero and receiver
volume infinite, in terms of initial and back pressures, pounds per square foot,
the volumes occupied by the fluid in both cylinders at cutoff, and at full
stroke in cubic feet. Dividing this by the volume of the lowpressure cylinder
V gives the mean effective pressure referred to the lowpressure cylinder,
from which the horsepower may be determined without considering the high
1 pressure cylinder at all. Hence, in the same units as are used for P b and P g ,
F*r F F/ F/l
(M.E.P. referred to L.P.) =Pt,~\ 2+loge ^+loge ^^ P 9 .
Vol Fft Ye Ye]
(300)
Proceeding as was done for simple engines, the work per cubic feet of fluid
supplied is found by dividing Eq. (299) by the volume admitted to the high
ipressure cylinder F/,, whence,
Workpercu.ft. supplied = p2+log e +log e P. . . (301)
Also applying the consumption law with respect to horsepower,
13 750 F
:Cu.ft. supplied per hour per I.H.P. = ^ to , ow) ^ ..... (302)
Lbs. supplied per hr. per I.H.P. = (m g p * Ki fo lqw) pfri ..... (303)
258
ENGINEERING THERMODYNAMICS
These last five equations, (299), (300), (301), (302), (303), while character
istic, are not convenient for general use in their present form, but are ren
( rel.pr. ) L
( atn>.pr.>
Q(bk.pr.)
H.P.
L.P.
IN DICATOR CARDS OF
H.P
? Cbk.pr.)
7 (rel.pr.) L
(atni.pr.)
L.P.
INDICATOR CARDS OF
EQUAL BASE AND
HEIGHT
H V
FIG. 82. Work of Expansive Fluid in Compound Engine with Infinite Receiver, Zero
Clearance. Cycle V, Logarithmic Expansion and Cycle VI Exponential Expansion.
dered so by substituting general symbols for initial and back pressures
displacement, cutoff, and amount of expansion for each cylinder.
WORK OF PISTON ENGINES
259
.",et (in.pr.) = initial or supply pressure, pounds per square inch = TA;
144
1 ' (rel.pr.)# = release pressure, in H.P. cylinder pounds per square inch =
p
p
(rel.pr.)z,= release pressure in L.P. cylinder, pounds per square inch= ~ ;
(rec.pr.) = receiver pressure, pounds per square inch=:^ = :rA;
p
(bk.pr.) = back pressure, pounds per square inch=Y~;
RH = ratio of expansion in highpressure cylinder = ^\
' b
R L = ratio of expansion in lowpressure cylinder \
R v = ratio of expansion for whole expansion =
Vt>
D H = displacement of highpressure cylinder = V d = V c ;
1 D L = displacement of lowpressure cylinder =V f =V g ;
1 R c = cylinder ratio = =f = =/ ;
DH Vd
' Z H = fraction of displacement completed up to cutoff in highpressure
' cylinder, so that Z H D H =V b = ^;
tlH
i Z L = fraction of displacement completed up to cutoff in lowpressure
cylinder, so that Z L D L =Ve=p'
KL
ibstitution of these general symbols in Eqs. (299), (300), (301), (302), and
(303) gives another set of five equations in useful form for direct substitu
tion of ordinary data as follows :
Work of cycle
. (304)
W ]^r  144(bk.pr.)l>L (a)
= 144A, I (in.pr.)^(2+log, y +lo&  ~^V) ~ ( bk P r 
Jtic\ AH AL, HC&L/
(in.pr.)^^l 2+loge RH+IO& X L ~]  (bk.pr.) j. (c)
(m.e.p.) Ibs. per sq.in. referred to L.P. cyl.
= (in.pr .
= (in.pr .)~(2+loge ^+log e sr  nrpJ ~ (bk.pr.) (b)
'#A ^// *L ZLKC/ J
. (305)
260
ENGINEERING THERMODYNAMICS
Work per cu.ft. supplied
= 144 T (in.pr.) (2+loge/i
[/ 1 1
(in.pr.) ( 2+logc = +loge =r
\ H L
Cu.ft. supplied per hr. per I.H.P.
13,750
(a)
(306)
(m.e.p. ref. to L.P.) R H Rc
13,750 Z H
(a)
(307)
J
(m.e.p. ref. to L.P.) R c
From this, of course, the weight in pounds supplied per I.H.P. results directly
from multiplication by the density of the fluid.
To these characteristic equations for evaluating work, mean pressure,
economy and consumption in terms of the initial and final pressures and cylin
der dimensions there may be added a series defining certain other general rela
tions of value in fixing the cycle for given dimensions and initial and final
pressures, and in predicting dimensions for specified total work to be done and
its division between high and lowpressure cylinders.
Returning to the use of diagram points and translating into the general
symbols as each expression is derived, there results,
Receiver pressure = P d = P e = Pb
(in.pr.)^% (6)
. . . (308)
Highpressure cylinder release pressure = P c = ,
.'. (rel.pr.)* = (in.pr.)
KM
= (in.pr.)Z*
y
Lowpressure cylinder release pressure = P f =P e ~.
i
.'. (rel.pr.) L = (in.pr.)
R v
(rol.pr.)w
Rc
(a)
(6)
(a)
(6)
(c)
(d)
. (309)
(310)
WORK OF PISTON ENGINES
261
Division of work between cylinders may be made anything for a given load by
suitably proportioning cylinders, and equations giving the necessary relations
to be fulfilled can be set down. It is quite common for designers to fix on
equal division of work for the most commonly recurring or average load or that
corresponding to some high pressure cutoff or lowpressure terminal pressure,
generally the latter. Therefore, a general expression for dimensional rela
tions to be fulfilled for equal division of work is useful. On the other hand,
for an engine the dimensions of which are determined, it is often necessary to
find the work division for the imposed conditions, so that the following equa
tions are of value.
From Eqs. (296) and (297), noting that P d = P e = P &1 /,
Highpressure cylinder work _
Lowpressure cylinder work ~
PtF
1+log> ^ _
1+loge RH~^
tic
(a)
1 1
1+loge ~
. . (311)
This is a general expression for work division between the cylinders in
terms of (a) ratio of expansion in each cylinder, initial and backpressure
ratio and cylinder ratio, or, in terms of (6) cutoff in each, associated with cylinder
and pressure ratios.
This expression Eq. (311) is less frequently used in its general form as above,
than in special forms in which the work of the two cylinders in made equal,
or the expression made equal to unity. The conditions thus found for equal
division of work between cylinders may be expressed either (a) in terms of initial
and back pressures, release pressure of lowpressure cylinder and ratio of L.P.
admission volume to H.P. displacement, and cylinder ratio, or (6) cutoff in
high and lowpressure cylinders, initial and back pressures and cylinder ratio.
Still more special conditions giving equality of work may be found (c) when the
262 ENGINEERING THERMODYNAMICS
cylinder ratio is made such that equality of work is obtained at all loads by
equalizing high and low cutoffs.
(a) To find the first set of conditions, equate Eqs. (296) and (297) from the
first part of this section, and by simplification there results,
or
~ ~ ^V e
Introducing the usual symbols and putting in addition
Lowpressure admission volume _ Ve _ _ _ (rel.pr.)//
Highpressure displacement volume V c (rec.pr.)
Therefore,
&?L>^_11
(rel.pr.W x J
[ L(rel.pr.) L xj (in.pr.) 1*
Rc= \e ; , N x\ .
(rel.pr.) L J
This is of value when a given release pressure is to be reached in the low
pressure cylinder and with a particular value of lowpressure cutoff volume
as fixed by x in terms of highpressure cylinder displacement.
(b) Again for equal division of work, make Eq. (311) equal to unity,
whence,
1+Io& _
ZIH KC&L
or
which may be reduced to the following, solving for R c ,
^ = ~^j^^T ~' ' ' (318)
Z H (in.pr.)
Equal division of work for given initial and back pressure is to be obtained
by satisfying these complex relations Eq. (313) between the two cutoffs, or their
equivalent ratios of expansion in connection with a given cylinder ratio, or the
relation between pressures and volumes in Eq. (312) equally complex.
WORK OF PISTON ENGINES
263
(c) An assumption of equal cutoff in both cylinders gives results which are
of .interest and practical value, although it is a special case. Eq. (313) then
becomes, when ZH ZL or R H = RL,
As would be expected, this may also be derived from Eq. (312), since
1 bk.pr. (rel.pr.)z,
=, , v and x= ,, . ^ \ under these conditions.
x (rel.pr.)z, (bk.pr.)
The receiver'pressure under these conditions is constant, and is, from Eq. (308)
(in.pr.) (in.pr.)
/ r> / \
KG /in.pr. y
\bk.pr./
.(315)
The highpressure release pressure is not affected by any change in the low
pressure cutoff, and hence Eq. (309) gives the value of highpressure release,
pressure for the case. Lowpressure release pressure Eq. (310) may be expressed
for the case of equal cutoff,
/ r \*7 V ~\ i '
(tel.pr.)i = y^^^ = ZJ (in.pr.) (bk.pr.) (a) . .
= ^ (in.pr.) (bk.pr.)
tin\~
(6) . . . .
(316)
The foregoing equations up to and including Eq. (311), are perfectly general,
and take special forms for special conditions, the most important of which is
that of complete expansion in both cylinders, the equations of condition for which
are, referring to Fig. 82.
Pf=p ,
which, when fulfilled, yield the diagram, Fig. 83. These equations of condi
tion are equivalent to fixing a relation between the cutoff in both the high
and lowpressure cylinders, and the volume of highpressure cylinder with respect
to the lowpressure volume, so that
Pf
V b = Vf ~, or symbolically,
*b
^/bJc.prA /bk.prA '
Da \rn.pr. / \in.pr. /
R " = R~ c \tiS^) {b
(317)
264
ENGINEERING THERMODYNAMICS
Similarly the lowpressure cylinder cutoff volume must equal the high
pressure displacement volume or D H =
DM 1
(6)
(318)
indicating that lowpressure cutoff is the reciprocal of the cylinder ratio. Making
the necessary substitution there result the following equations for this cycle
which, it must be noted, is that for most economical use of fluid in compound
* z
Z L D r
(i
IT.)
( r c.pr.)
N C
E(rel,pr.)
FIG. 83. Special Case of Cycles V and VI, Complete Expansion in both Cylinders of
Compound Engine with Infinite Receiver and Zero Clearance.
cylinders without clearance and with infinite receiver, and in which the same
work is done as in Cycle I, for simple engines at best cutoff.
From Eq. (308)
. . . (319)
From Eq. (309),
WOEK OF PISTON ENGINES
265
From Eq. (310),
x (in.pr.) (in.pr.) ,, . x
.pr.)z, = Vp = i / ' x = (bk.pr.)
KcKH yin.pr.;
(321)
p _
Kc
Rc (bk.pr.)
From Eq. (311),
High
pressure
cylinder
work
Re
loge RH
(a)
/1A
Low
pressure
cylinder
work .
og^ga^w
loge RL
1
. (322)
For the case of most economical operation, that of complete and perfect
expansion in both cylinders, there may be set down the four characteristic
Eqs. (304), (305), (306), (307) with suitable modifications to meet the case.
These become
Work of cycle = 144(in.pr.)Dzr
.pr
/in.pr. \
Vbk.pr./
bk.pr
^Z. . . (323)
. ,(324)
R
*
1 Work per cu.ft. supplied = 144(in.pr.) log e ~ = 144(in.pr.) log e R v . (325)
. . . . (226)
13,750 /bk.prA
Cu.ft. per hr. per I.H.P. = 7  , . T P X(T ) (a
(m.e.p. ref. to L.P.) \in.pr. /
13,750
(m.e.p.ref . to L.P.) Rv
(b)
ijor .equal division of work with complete expansion in both cylinders, the
ratios of Eqs. (317) and (318) becomes
bk.pr A *
(327)
id this is evidently a case to which Eqs. (314) and (315) apply without change.
266 ENGINEERING THERMODYNAMICS
Example. 1. Method of calculating diagram, Fig, 82.
Assumed data for Case A:
p a = p b = 100 Ibs. per sq.in. abs. V a = V n = V m = cu.ft.
p n= p d= p e= 50 Ibs. per sq.in. abs. V c = V d = .6 cu.ft.
p m =p g = 10 Ibs. per sq.in. abs. V f = V g = 2 cu.ft.
P c = 60 t lbs. per sq.in. abs. V e = .8 cu.ft.
To obtain point B:
p ar\
Vb = VcX ^. = ,Q X =.36cu.ft.
* b
To obtain point F:
=201bs. persq.in.
Vf
To construct the indicator cards:
Lay off ND of the PV diagrams to equal the length of the card, and NA perpen
dicular to it at N to equal the height of the card. Cut off equals AB+ND. From
on card lay off this ratio times the length of the card. From D on the card lay off
a perpendicular equal to CD of the PV diagram reduced by the same proportioi
as AN of the card is to AN of the diagram. Join the points B and C by a curv
through points located from intermediate points on the PV diagram. The low
pressure card is constructed in same manner.
Example. 2. A 12 and 18x24in. steam engine without clearance runs on 15i
Ibs. per square inch absolute initial pressure, 10 Ibs. per square inch absolute back pressure
and has a speed of 125 R.P.M. What will be (a) the horsepower for cutoff in H.P
cylinder, (b) pounds of steam per I. H.P. hour, (c) terminal pressures, (d) L.P. cutoff
for continuous expansion, (e) work done in each cylinder.
NOTE: B for 150 Ibs. =.332.
(a) From Eq. (305)
(m.e.p.) referred to L.P, cylinder is
(in.pr.) I 2+log e J?//+log e #z, ] (bk.pr.).
KnKc \ KC/
In this case
I) =2.25, ^=2.25,
since vol. of L.P. cyl. at cutoff must be equal to the entire volume of the high fo
continuous expansion, hence
(m.e.p.)=150XX(2+.69+.81l)10=73.3 Ib. sq. inch,
22
and
(b) From Eq. (307)
WORK OF PISTON ENGINES 267
(c) From Eq. (309)
(rel.pr.)# = (in.pr.)^a,
= 150X^=75 Ibs. sq.in,
and from Eq. (310) we have
75
= =33.3 Ibs. sq.in.
(e) From Eq. (311)
1 1
H.P. work ZH RcZL
L.P. work (bk.pr.) R c '
1+.69
2.25 X. 44 .69 , Bfl
 = .456,
81  L51
H.P.work =. 456 XL.P. work,
f  H.P. work+L.P. work =282 I.H.P.
ce
H.P. work =88 I.H.P.
L.P. work = 194 I.H.P.
Prob. 1. What must be the cylinder diameters of a cross compound engine to run
100 Ibs. per square inch absolute steam pressure, 18 ins. of mercury vacuum and to
Develop 150 H.P. at a speed of 200 R.P.M. with \ cutoff in each cylinder, if cylinder
mtio is 3 and stroke is 18 ins.? Engine is doubleacting and assumed to have no
clearance.
Prob. 2. What will be the release pressure in each cylinder and the receiver
aressure of the engine of Prob. 1? If cutoff were reduced to  in H.P. cylinder,
few would these pressures be affected and to what extent? How would the horse
power change?
Prob. 3. A 15 and 22 X30in. infinite receiver engine has no clearance, a speed of
J50 R.P.M., initial pressure 125 Ibs. per square inch gage. What will be the horse
iower and steam consumption for a H.P. cutoff of J, i, f , \, and that value which
fill give complete expansion in highpressure cylinder? Lowpressure cutoff to be
ixed at \.
 NOTE: $ for 150 Ibs. gage = .363.
268 ENGINEERING THERMODYNAMICS
Prob. 4. What will be the release and receiver pressures, and the work done in each
cylinder for Prob. 3?
Prob. 5. An 18 and 24x30in. infinite receiver engine is to be operated so as
to give complete expansion in both cylinders. What will be the cutoff to accomplish
this and what horsepower will result if the initial pressure is 100 Ibs. and back pressure
10 Ibs. per square inch absolute?
Prob. 6. Draw the PV diagram for following cases. Cylinder ratio 1 to 2.5,
(in.pr.), 100 Ibs. per square inch absolute, (bk.pr.), 20 Ibs. per square inch absolute,
H.P. cutoff (a)=l (6)=J, (c)=i L.P. cutoff (a) =4, (6) =&, (c) =1 .
Prob. 7. For the following conditions find the horsepower, steam used per hour,
receiver pressure and release pressures. Engine, 10 and 15x24in. 150 R.P.M.,
125 Ibs. per square inch gage initial pressure, 2 Ibs. per square inch absolute, back
pressure, \ cutoff in highpressure cylinder, M> cutin lowpressure cylinder with
infinite receiver.
NOTE: 8 for 125 Ibs. =.311.
Prob. 8. An infinite receiver engine is to develop 150 H.P. at 200 R.P.M. when
initial pressure is 150 Ibs. per square inch absolute. Cylinder ratio is 1 to 3 and
back pressure is one atmosphere. What must be its size if the stroke is equal to
the lowpressure cylinder diameter for \ cutoff in the highpressure cylinder and
 cutoff in the lowpressure cylinder?
Prob. 9. Find by trial the cutoffs at which work division will be equal for an
infinite receiver engine with a cylinder ratio of 2.5, an initial pressure of 100 Ibs. pei
square inch absolute and a back pressure of 5 Ibs. per square inch absolute?
10. Compound Engine with Infinite Receiver, Exponential Law. Nc
Clearance, Cycle VI. General Relations between Pressures, Dimensions
and Work. Again referring to Fig. 82, which may be used to represent thi:
cycle also, the work of each cylinder may be expressed as follows, by th
assistance of Eq. (254) derived in Section 4.
. . . (3281
. . . (3291
where Z H is the cutoff in the high pressure, = ~ and Z L , lowpressure cuto:
V e
= Y' In combining these into a single equation for the total work, the terr
for receiver pressure (rec.pr.) should be eliminated. Referring to Fig. 82,
(*.)P.P.P$)'<tor.)()', . . .
hence
WORK OF PISTON ENGINES 269
a rather complex expression which permits of little simplification, but offers
no particular difficulty in solution.
Mean effective pressure referred to the lowpressure cylinder is
<*>
Work per cubic feet fluid supplied may be found by dividing Eq. (331) by
the supply volume, which in terms of lowpressure displacement is
(Sup.Vol.)=D^ (333)
The consumption of fluid, cubic feet per hour per indicated horsepower is
jConsumption cu.ft. per hr. per I.H.P. = ; ' ^, . . . (334)
which is the same expression as for the logarithmic law. Multiplying this by
81, the initial density of the fluid, pounds per cubic feet, gives consumption,
^pounds per fluid hour per I.H.P.
S ! The receiver pressure has already been determined in Eq. (330).
[isj Release pressure of the highpressure cylinder is
(rel.pr.)*=(in.pr.)Z* s , . . V ; V ! . ; . . (335)
for the lowpressure cylinder,
(rel.pr.)z,= (in.pr.)(?Y (a)
(336)
(m.pr.)
rhere R v is the ratio of maximum volume in the low pressure, to, volume at
r>
jutoff in the high, and equals =.
AH
The distrubition of work between the high and lowpressure cylinders may
>e found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by
{leans of Eq. (330)
sl / \RcZj
TT7 r / r7 \ / ~ r/ a \ \ /U1r * \ 1 " * * v ' /
W L r
270
ENGINEERING THERMODYNAMICS
Equality of work in the two cylinders will be obtained if this expression
is equal to unity, giving a complex relation between high and lowpressure
cutoffs, cylinder ratio and ratio of initial and back pressures, to be satisfied.
It is found at once that the simple conditions for equality in the case of logarith
mic law will not give equality of work for the exponential law. There is, how
ever, a case under this law which yields itself to analysis, that of complete expan
sion in both cylinders, without overexpansion. The conditions for equality
of work for this case will be treated after deriving work and mean effective
pressure for it.
Complete expansion, without overexpansion, in both cylinders may be
represented by Fig. 83.
AB
NC
and since
P =___
DH NC Z L .
The true ratio of expansion = J R F =^ = 1 =^ J but this is also equal to
AB AH&L AH
\F ^r") * dUe t0 the laW f th
By means of Eq. (257) in Section (4) the work of the two cylinders may be
evaluated,
(a)
(338),
but since
(339
I
WORK OF PISTON ENGINES
271
The total work is evidently the same as that of a cylinder equal in size to
i?
the lowpressure cylinder with a cutoff equal to ~, working between the
He
given (in.pr.) and (bk.pr.) and may be stated by reference to Eq. (257),
Section 4, or by taking the sum of W H and W L given above,
which reduces to
Z H s
'Rcsl
{ fc
(340)
For this case of complete expansion in both cylinders, the ratio of high to
Howpressure work is given by division and cancellation,
TF
W
L ( V
\Rc)
f Re"
AH/
Rc^l '
(341)
Equality of work, obtained by placing this expression equal to unity, pro
ivides the condition that
sl
*
Re
equal work and complete expansion, and
sl
Z H =
bk.
.pr s \n.pr.
= ~
1
:l
(342)
(343)
Since Z L = for complete expansion, 'and (in Fig. 83) P c V c s = PfV f s , the
iiver pressure, P c , is
(rec.pr.) = (bk.pr.)(^y=(bk.pr.)fl<7', !? . , ',. . . (344)
which Re will have the value given above if work is equally distributed.
Example 1. What will be (a) the horsepower, (6) consumption, (c) work ratio,
receiver and release pressures for the following conditions? Engine 12 and
X24 ins., running at 125 R.P.M. on initial air pressure of 150 Ibs. per square
absolute, and back pressure of 10 Ibs. per square inch absolute, with \ .cutoff
highpressure cylinder and continuous expansion in lowpressure cylinder. Exponent
expansion curve = 1.4 for compressed air, infinite receiver.
272 ENGINEERING THERMODYNAMICS
(a) From Eq. (332)
which, on substituting values from above, gives for (m.e.p.) 63 Ibs. per sq. inch.
Hence, the indicated horsepower =242.
(6) From Eq. (334)
1 Q '7 f\C\ *7
Compressed air per hour per I.H.P. =  = cu.ft.,
m.e.p. R c
which, on substitution, gives
13,750 .5
(c) From Eq. (337)
W
H
W L J z /JkV/l^ 8 '^ /bk.pr.
which gives
5f * i / O
=.294.
\ 10 1
and
Wj/ + PTi=242I.H.P.
Hence
WW=56 I.H.P.
and
W L = 184 I.H.P.
(d) From Eq. (330)
(rec.pr.) = (in.pr.) ( ^ Y = ISO/ ^ r \ 1 ' 4 =57 Ibs. per sq.in.
From Eq. (335)
(rel.pr.)/, = (in.pr.)Z^ s ,
150x(.5) 14 =571bs. persq.in.
WOEK OF PISTON ENGINES 273
From Eq. (336)
= 150 +21.85=6.85 Ibs. per sq.in.
These values may be compared with those of Ex. 1, Section 9, which were for the
same data with logarithmic expansion.
Prob. 1. What will be the horsepower and steam used per hour by the follow
.ing engine under the conditions given? Cylinders 18 and 30x48 ins., speed 100
R.P.M., initial pressure 150 Ibs. per square inch absolute, backpressure 10 Ibs. per
square inch absolute, steam continually dry. Cutoff at first in highpressure and
\ in low, and then ^ in each infinite receiver.
Prob. 2. The very large receiver of a compound pumping engine is fitted with safety
valve which is to be set to blow at 25 per cent above ordinary pressure. The cylinder
ratio is 1 to 3.5, and cutoffs are f in high and  in low. If initial pressure is 125
Ibs. per square inch gage, for what must valve be set? What vacuum must be carried
in the condenser to have complete expansion in lowpressure cylinder? Superheated
steam.
Prob. 3. A compound engine is to be designed to work on superheated steam of
125 Ibs. per square inch absolute, initial pressure and on an 18inch vacuum. The
load which it is to carry is 150 horsepower and piston speed is to be 500 ft. per
minute at 200 R.P.M. Load is to be equally divided between cylinders and there is
to be complete expansion in both cylinders. What must be cylinder sizes, and
what cutoffs will be used for an infinite receiver?
Prob. 4 V How will the economy of the two following engines compare? Each is
14 and 20x24 ins., runs at 200 R.P.M. , on compressed air of 100 Ibs. per square
inch gage pressure, and 15 Ibs. per square inch absolute exhaust pressure. Lowpres
sure cutoff of each is ^ and high pressure of one is I, the other, I. Infinite
receivers.
Prob. 5. A compound engine 12 and 18x24 ins. is running at 200 R.P.M. on
superheated steam of 100 Ibs. per square inch absolute pressure and exhausting to a
condenser in which pressure is 10 Ibs. per square inch absolute. The cutoff is I in
highpressure cylinder and  in lowpressure cylinder. Compare the power and steam
consumption under this condition with corresponding values for wet steam under
same conditions of pressure and cutoff and infinite receivers.
Prob. 6. The initial pressure of an engine is 150 Ibs. per square inch absolute, the
back pressure one atmosphere, the cylinder ratio 3. As operated, both cutoffs are at .
What will be the receiver pressure, highpressure release pressure, and lowpressure
release pressure? What will be the new values of each if (a) highpressure cutoff is
made i, (6) I, without change of anything else, (c) low pressure cutoff is made i,
(d) f, without change of anything else? Infinite receiver, s = 1.3.
Prob. 7. In the above problem for cutoff in each cylinder how will the release and
receiver pressures change if (a) initial pressure is raised 25 per cent, (6) lowered 25
per cent, (c) back pressure raised 25 per cent, (d) lowered 25 per cent?
Prob. 8. How many pounds of initially dry steam per hour will be required to
supply an 18in. and 24x30in. engine running at \ cutoff in each cylinder if speed
274
ENGINEERING THERMODYNAMICS
be 200 R.P.M., initial pressure 100 Ibs. per square inch gage and back pressure 5
Ibs. per square inch absolute? Expansion to be adiabatic and receiver infinite.
NOTE: d for 100 Ibs. =.26.
^ Sup. Vol.
INCOMPLE' [EXPANSION
OVER EXPANSION V
FIG. 84. Work of Expansive Fluid in Compound Engines with Finite Receiver, Zero Clear
ance. Cycle VII, Logarithmic Expansion; Cycle VIII, Exponential Expansion.
11. Compound Engine with Finite Receiver. Logarithmic Law. No
Clearance, Cycle VII. General Relations between Dimensions and Work
when H.P. Exhaust and L.P. Admission are not Coincident. The diagrams,
Fig. 84, while showing only two degrees of expansion, that of over and under
in both cylinders, suffice for the derivation of equations applicable to all
degrees in either cylinder. Volumes measured from the axis AL are those
WORK OF PISTON ENGINES
27,")
occupied by the fluid in either cylinder alone, while fluid volumes entirely in the
receiver, or partly in receiver, and in either cylinder at the same time are meas
ured from the axis A'L'. No confusion will result if all volumes represented by
points be designated by the (V) with a subscript, and to these a constant rep
resenting the receiver volume be added when part of the fluid is in the receiver.
Then,
Highpressure cylinder work is
(345)
Lowpressure cylinder work is
W L = P n O\og e
Total work
+P,Felog,^P f F f . . . . (346)
P ( ,V . . (347)
These expressions include some terms not known as initial data and may
be reduced by the following relations,
and
Hence
Dividing by the lowpressure cylinder displacement, V ff , the result will be
the mean effective pressure referred to the lowpressure cylinder,
(M.E.P. ref. to L.P.)
276
ENGINEERING THERMODYNAMICS
A similar division but with the volume supplied,
Work per cu.ft. supplied
as the divisor, gives
Also as in previous cases
Cu.ft. supplied per hr. per I.H.P.
(m.e.p. ref. to L.P.)
(351)
Of course, the weight per hour per I.H.P. follows from Eq. (351) by introduc
ing the density as a multiplier.
While the last four equations can be used for the solution of problems, it is
much better to transform them by introducing dimensional relations as in the
previous cases developed.
Let (rec.pr.)i= maximum receiver pressure P n , which is also the initial admis
sion pressure for the lowpressure cylinder;
(rec.pr.) 2 = minimum receiver pressure P e , which is the terminal admission
pressure for the lowpressure cylinder and that at which
expansion begins there;
"
receiver volume
highpressure cyl. displ.
____ H
" ~ ' l '
Other symbols necessary are unchanged from the meaning imposed in Section (9).
Substitution in Eqs. (348), (349), (350), and (351) gives the following set
in a form for direct substitution of ordinary data:
Work of cycle
= U4(m.pr.)Z H D H 1 +log e +log e 
 (35)
log, (l+I
 144(bk.pr.)Z>z, (6)
WORK OF PISTON ENGINES
277
(m.e.p. ref. to L.P.)
= (m.pr.)f^ 1 1 +loge 
tic

= (in.pr.)
Work per cu.ft. supplied
= 144(in.pr.)(l+log e ^+log e
+ 1 +
144(in.pr.) l+log e
R L +
Cu.ft. supplied per hr. per I.H.P.
13,750
(m.e.p. ref. to L.P.) R c
13,750 1
X
(m.e.p. ref. to L.P.) R H Rc
(6)
(353)
pr.)^ W
(6)
. (354)
xf* (a)
(355)
It is desirable at this point to introduce a series of expressions fixing the
relations between the dimensions, the cycle that may follow, and the fluctua
tions in the receiver pressure, and for the selection of cylinder and receiver
dimensions for a required output of work and division of it between cylinders.
In doing this it will be convenient to start with diagram points and finally
substitute general symbols in each case. There will first be established the
maximum and minimum receiver pressures and the fluctuations.
Maximum receiver pressure
=P IZe+o) = /py\ (Ve+ o\ _ p /T. , r,\
. (356)
278 ENGINEERING THERMODYNAMICS
Minimum receiver pressure
(a)
/^
(357)
(in.pr.)
Fluctuation in receiver] pressure = (P n Pe)=Pb^>
Z D Z
.'. (rec.pr.)i (rec.pr.) 2 = (in.pr.) * " = (in.pr.)j (a)
(358)
It is interesting to note that the minimum receiver pressure is exactly the same
as the value of the constantreceiver pressure for infinite receiver, so that limit
ing the size of receiver does not affect the point E, but only raises point N higher,
tending to throw more work on the L.P. cylinder for the same valve setting.
The two release pressures P c and P/ can be evaluated as in the case of the
infinite receiver, as both these points lie on the common expansion line, which
is not at all affected by the receiverpressure changes, and the values are the
same as for the infinite receiver, and are here reproduced from Eqs. (309) and
(310) with new numbers to make the set of equations complete:
(rel.pr.)//=(in.pr.)~ (a)
KH
= (in.prr)Z* (6)
(rel.pr.)z, = (in.pr.)  (a)
(359)
(360)
where R v is the ratio of maximum volume in the low to the volume at cutoff
in the highpressure cylinder.
Division of work between the cylinders cannot, as pointed out, be the same
as for the infinite receiver, the tendency being to throw more work on the low
as the receiver becomes smaller, assuming the cutoff to remain the same. As,
therefore, equal division was obtainable in the case of infinite receiver with
WORK OF PISTON ENGINES
279
equal cutoffs when the cylinder ratio was equal to the square root of initial
over back pressure, it is evident that a finite receiver will require unequal cut
offs. As increase of lowpressure admission period or cutoff fraction lowers
the receiver pressure and reduces the lowpressure work, it follows that with
the finite receiver the lowpressure cutoff must be greater than the high for
equal work division, and it is interesting to examine by analysis the ratio
between them to determine if it should be constant or variable.
For equal work division Eqs. (345) and (346) should be equal, hence by
diagram points
P,F
(l+log, ) 
r =PJ> 10g e
+ P.V. log, 
P
log,
hence for equal division of work, the following relations must be satisfied :
(log, R a 
= 1 +
Re
/
\
bk.pr.
in.pr.
R H Rcl. (361)
It will be shown later that when expansion is complete in both cylinders
and work equal that the highpressure cylinder cutoff or the equivalent ratio
of expansion bears a constant relation to that of the low, according to
^ = 3, . (362)
KL
in which a is a constant depending only on the size of the receiver. It will
also be shown that the cylinder ratio is a constant function of the initial and
back pressures and the receiver volume for equal division of work, according to
:^V, (363)
in which (a) is the same constant as in Eq. (362). It is impo tant to know if
these same values will also give equal division for this general case. Substi
tuting them in Eq. (361)
280
ENGINEERING THERMODYNAMICS
Here there is only one variable, R L , the evaluation of which can be made by
inspection, for if
the equation will become
2 log, a = (l+y)
or
which is a constant.
1.001
RECEIVER VOLUME EQUALS
2/X H.P. DISPLACEMENT
(364)
.40
High Pressure Cut Off
,80
.100
FIG. 85. Diagram to Show Relation of High and LowPressure Cutoffs for Equal Work in
the Two Cylinders of a Finitereceiver Compound Engine with Zero Clearance
and Logarithmic Law.
As only one constant value of lowpressure ratio of expansion or cutoff
satisfies the equation for equal division of work when there is a fixed ratio
between the values for high and low, that necessary for equal division with
complete expansion in both, it is evident that equal division of work between
the two cylinders cannot be maintained at all values of cutoff by fixing the
ratio between them. As the relation between these cutoffs is a matter of some
interest and as it cannot be derived by a solution of the general equation it is
given by the curve, Fig. 85, to scale, the points of which were calculated.
WORK OF PISTON ENGINES
281
A special case of this cycle of sufficient importance to warrant derivation
of equations because of the simplicity of their form and consequent value
in estimating when exact solutions of a particular problem are impossible,
is the case of complete and perfect expansion in both cylinders. For it the
following equations of condition hold, referring to Fig. 84,
il
INDICATOR CARDS OF EQUAL
BASE AND HEIGHT
!H
IG. 86. Special Case of Cycles VII and VIII Complete Expansion, in both Cylinders of the
Finite Receiver Compound Engine. Zero Clearance.
which when fulfilled yield the diagram, Fig. 86. These equations of conditions
are equivalent to fixing the cutoff in both high and lowpressure cylinders,
and the volume of the high with respect to the lowpressure volume . Accordingly,
(365)
Also for the lowpressure cylinder the cutoff volume must equal the whole
lighpressure volume, or D H = Z L D L . Therefore,
= ' (a)
He
(366)
2X12
ENGINEERING THERMODYNAMICS
Substituting these equations of condition in the characteristic set Eqs. (352),
(353), (354), and (355), there results the following for most economical operation:
Work of cycle
W
log,
144(bk.pr.)I> i
(a)
144(in.pr.)Z)z, ^ = 144(in.pr.)D i
/in.pr. \
Vbk.prJ
R
(367)
(m.e.p. ref. to L.P.) =
i /in.pr. \
/in.pr. \ ,. ,. ge \bk.pr./ ,. ,
(bk.pr.) log, ^ J = (m.pr.) . = (m.pr.)
e  /0 ^o\
 . (368)
W
Work per cu.ft. supplied = ^fr
(bk.prODz, . /in.pr. \ .... V1 /in.pr.
144 , , loge , , =144(m.pr.) lofotri
p /bk.pr A 5 \bk.pr. / 5 Vbk.pr.
J \in.pr. / H
. (369
Cu.ft. supplied per hr. per I.H.P.
13,750 x f bk P r A 13,750 1 ,
(m.e.p. ref. to L. P.) * \in.pr. / (m.e.p. ref. to L. P.) *R V '
For this special case of best economy the receiver and release pressures
of course, have special values obtained by substituting the equations c ,
condition Eqs. (365) and (366), in Eqs. (356), (357), (358), (359), (360).
(recp,), = (in.pr.) + = (in.pr.) +
WOKK OF PISTON ENGINES
283
Therefore
r>
.pr.)~;
(373)
(374)
(375)
These last two expressions might have been set down at once, but are
orked out as checks on the previous equations.
For equal division of work in this special case the general Eq. (361) becomes
(rec.pr.)i(rec.pr.) 2
(rel.pr.)*=(in.pr.) =(bk.prO#c=(rec.pr.) 2 ;
K H
Therefore
(376)
'his term, a, has already been used in previous discussions of equality of
prk, while the derivation of its value has not been made up to this point.
This indicates that ratio of cutoffs or individual ratios of expansion is a
inction of the receiver size for equal division of work.
From Eq. (376) the cylinder ratio can be found in terms of a, and the ratio
expansion. Referring to Fig. 86,
2
V e
ice the cylinder ratio is equal to a constant depending on the receiver size,
'Ultiplied by the value for the infinite receiver, i.e., the square root of the initial
by back pressure.
284 ENGINEERING THERMODYNAMICS
The highpressure cylinder ratio of expansion is
. . . (378)
and the corresponding value for the lowpressure cylinder is
V, 1 Vl /5nr7
, . . . (379)
For convenience in calculation Table, XII of values of a and a 2 is added foi
various size of receivers.
TABLE XII.
Receiver Vol.
[(l+y)log e (l + ^)l]
a 2
H.P. Cyl. Disp. l
.5
1.915
2.64
.75
1.624
3.67
1.0
1.474
2.17
1.5
1.322
.75
2.0
1.243
.55
2.5
1.198
.437
3.0
1.164
.359
4.0
1.1223
.262
5.0
1.0973
1.204
7.0
1.0690
1.143
10.0
1.0478
1.098
14.0
1.0366
1.068
20.0
1.0228
1.046
Infinite
1.0
1.0
At the end of this chapter there is presented a chart which gives the relatio]
between cylinder and receiver volumes, cylinder ratio, and high and lowpressur
cutoffs graphically.
The corresponding values of maximum and minimum receiver pressun
for equal division of work for this case of best economy are
(recpr.),  (rec.pr.) 2 = V ^^'\
(382
WORK OF PISTON ENGINES 285
Example 1. Method of calculating Diagram, Fig. 84.
Assumed data for case A:
p a =p b = 120 Ibs. per sq.in. abs. V a = V n = V m =0 cu.ft.
P m =P = 10 Ibs. per sq.in. abs. V* = A cu.ft.
V e = l cu.it.
= 1.2cu.ft.
V e = .Scu.it.
To find point (7:
.8
To obtain point E:
To obtain point D:
Pe(V e 
To obtain point N:
P e =P 6 = . =48 Ibs. per sq. inch.
Ve 1
48 v9 9
Pe(F,+0)=P d (Fc+0) or P d =^^=53 Ibs. per sq. inch.
48 v2 2
n )=P e (Fe+0) or P n =^=88 Ibs. per sq. inch.
To obtain point F:
48x1 . . ,
=24 Ibs. per sq. inch.
Example 2. Find (a) the horsepower, (6) steam used per hour, (c) the release
and receiver pressures for a 12 and 18x24in. engine with receiver twice as large as
the lowpressure cylinder when the initial pressure is 150 Ibs. per square inch absolute,
back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cutoffs  in
aiglipressure and such a value in the low pressure as to give complete expansion, and
learances zero.
(a) From Eq. (353)
jm.e.p.) = (in.pr.)  1 1 +loge RH flog* RL
which on substituting the above values gives
A. K\ / \\
10 = 73.3 Ibs.
le
nce I.H.P.=282.
(6) From Eq. (355) we have
fu.ft. steam per hour per horsepower = !   X p p =~~^ X =41.7,
(m.e.p.) Katie
286
ENGINEERING THERMODYNAMICS
(c) From Eqs. (356) and (357) for maximum and minimum receiver pressures
respectively :
+ R L \ and
RnRcl
maximum receiver pressure =
I 2 25
+ =91.5 Ibs. per sq. inch.
2.25
minimum
receiver pressure = 150 X^r~^= 75 Ibs. per sq. inch.
From Eqs. (359) and (360) for release pressures
(in.p,)Z and
high pressure cylinder release pressure = 1 50 X. 5 =75 Ibs., per sq. inch.
low pressure cylinder release pressure = =33.9 Ibs. per sq. inch.
''
These results may be compared with those of Example 1 of Sections 9 and 10, whicl
are derived for same engine, with data to fit the special cycle described in the particular]
section.
NOTE: In all the following problems clearance is to be neglected.
Prob. 1. A 12 and 18x24in. engine has a receiver equal to 5 times the volume of!
the highpressure cylinder. It is running on an initial pressure of 150 Ibs. per square
inch gage and exhausts to the atmosphere. It has a speed of 150 R.P.M. and the cutoffs
are & and } in high and lowpressure cylinders respectively. What is the horsepoweij
and the steam used in cubic feet per hour?
Prob. 2. What will be the release pressures, and variation of receiver pressure foil
an engine in which the cylinder ratio is 3, cutoffs  and , in high and low, initial pres
sure is 100 Ibs. per square inch absolute, and receiver 2 times lowpressure cylindei]
volume?
Prob. 3. Show whether or not the following engine will develope equal cylindei
work for the conditions given. Cylinder diameters, 15 and 22 in., initial pressure
Ibs. per square inch gage, back pressure 10 Ibs. per square inch absolute, cutoffs ]
and f , receiver volume 4 times highpressure cylinder, strokes equal.
Prob. 4. For the same conditions as above, what lowpressure cutoff would giv<
equal work?
Prob. 6. What will be the most economical load for a 16 and 24x30in. engii
running at 125 R.P.M. on 150 Ibs. per square inch absolute initial pressure and atj
mospheric backpressure? What will be the economy at this load?
Prob. 6. What will be the release and receiver pressures for the above engine i
the receiver has a volume of 15 cu.ft.?
WORK OF PISTON ENGINES 287
Prob. 7. Find the cutoffs and cylinder ratio for equal work division and complete
expansion when initial pressure is 150 Ibs. per square inch absolute and back
pressure is 10 Ibs. per square inch absolute, receiver four H.P. volumes.
Prob. 8. Will a 14 and 20x20in. engine, with a receiver volume equal to 5 tunes
:he, H.P. cylinder and running on I cutoff on the highpressure cylinder and I cutoff
MI the low, with steam pressure of 100 Ibs. per square inch gage and back pressure
)f 5 Ibs. per square inch absolute, have complete expansion and equal work distri
Dution? If not, what changes must be made in the cutoff or initial pressure?
Prob. 9. What must be the size of an engine to give 200 1. H.P. at 150 R.P.M. on an
nitial steam pressure of 150 Ibs. per square inch absolute, and 10 Ibs. per square inch
ibsolute back pressure, if the piston speed is limited to 450 ft. per minute and complete
expansion and equal work distribution is required? Receiver is to be 6 times the volume
rf highpressure cylinder and H.P. stroke equal to diameter.
12. Compound Engine with Finite Receiver. Exponential Law, No
Clearance. Cycle VIII. General Relations between Pressures, Dimensions,
and Work, when High Pressure Exhaust and Lowpressure Admission are
Independent. The diagram Fig. 84 may be used to represent this cycle, as well
as cycle VII, by conceiving a slight change in the slope of the expansion and
receiver lines. Using the same symbols as those of the preceding section,
and the expression for work as found in Section 7, Chapter I,
^r^kl^'] }
nd the last term in the equation for W H within the bracket may therefore be
written
Cm.pr.)Z, (y\ (Z,\ /_ y_ +l \ ^ pLY
~ l \y+l)

Rcz
2SS ENGINEERING THERMODYNAMICS
and hence by simplifying the first two terms also,
Work of the lowpressure cylinder may be expressed in terms of pressure
and volumes at N, E, and G, but it is convenient to use instead of the pressure
at N or at E, its equivalent in terms of the point B. The pressure at N is
and when multiplied by the receiver volume yDn, it becomes
At E the product of pressure and volume is
(rec.pr.) 2
Using these quantities, the following equation gives the work of the low
pressure cylinder:
and the total work is, by adding (W H ) and (W L ),
This Eq. (385) is the general expression for work of the zero clearance com
pound engine with exponential expansion, no clearance, and finite receiver,
From this the following expressions are derived :
(m.e.p. ref. to L.P.)
i
WORK OF PISTON ENGINES 289
Vork per cu.ft. supplied is
_144^I.Z a .'+^V~ 1 p + lY IYJLV 1
! 1 \v) \RcZ,7 V l\y+l/
.pr.). . (387)
}u.ft. supplied per hr. per I.H.P.
13,750
(m.e.p. ref. to L. P.) R c '
(388)
(389)
; .... (390)
*=(in.pr.)Zj/; ..... ..... (391)
V ..... (392)
If work is equally divided between the cylinders, W H , Eq. (383), and W L ,
(384), will become equal, hence
._,

.
(in.pr.) Z ff
This equation shows conditions to be fulfilled in order that an equal division
work may be obtained. It does not yield directly to a general solution.
When expansion is complete in both cylinders,
1 /bk.prA (Z H \>
^ and . ) = l 7^ ) .
Rc \in.pr. / \Rc/
Rc
Introducing these values in the general expression Eq. (385) for work of
jiis cycle, it may be reduced to the following:
..... (394)
From which are obtained
(m.e.p. ref. to L. P.) = (in.pr.)p
~ T [l^y" j
s f (Zn\'~ 1 ~\
r ork per cu.ft. supplied = 144(in.pr. 377 1~ l"p~ )
. (395)
(396)
290 ENGINEERING THERMODYNAMICS
13,750 _
Cu.ft. supplied per hr. per I.H.P. = (m e p ref to L p }
13,750
(WESLV. (397
. P.) \in.pr. /
(m.e.p. ref. to L
If work is equally divided and complete expansion is maintained in botl
cylinders Eq. (381) becomes
which may be simplified to the form,
s. ... (398
where Rv is the ratio of maximum lowpressure volume, to the highpressui
volume at cutoff,
hence
and the value of Rv may be found from original data,
ErMc L ) (39
Vbk.pr./
Eq. (398) may easily be solved for Z H , from which the required cylind
ratio may be found by,
Rc=Z H R v . ......... (40(
This is the cylinder ratio which gives equal work in the two cylinders am
complete expansion in both, when used with the value found for the higl
pressure cutoff Z H , the assumed initial and back pressures, and the assume
raio, y, of receiver volume to highpressure displacement.
WORK OF PISTON ENGINES
291
Example. Find (a) the horsepower, (6) steam used per hour, (c) the release
nd receiver pressures of a 12 and 18x24in. engine, with a receiver twice as large as
ic lowpressure cylinder when the initial pressure is 150 Ibs. per square inch absolute,
ack pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cutoffs in the
igh and such a value in the low as to give complete expansion. Exponent for ex
ansion curve = 1.4.
(a) From Eq. (386)
m.e.p.,
yhich, on substituting above values, gives
50 .5 [ . '_ / .5 \  4 / 4.5
'fence
(m.e.p.) =57.5 Ibs. per sq.in.,
I.H.P.=221.
(b) From Eq. (388)
r, , 13,750 Z H
Oubic reet 01 steam per hour per horsepower = =.
m.e.p. Re
13,750 .5
lence total pounds per hour will be
53.2 X221X. 332 =3910.
From Eqs. (389) to (392):
J Z H \ s (,RcZ L \ s
(rec.pr.)i = (m.pr.) () (l +) ,
(7 \ s
T I
(rel.pr.)^ = (m.pr.)Z H s ,
292 ENGINEERING THERMODYNAMICS
These, on substitution of the proper numerical values, become:
(g\ 1.4 / j \ 1.4
~j f Xl + ) =75 Ibs. per sq. inch,
(rec.pr.) 2 = 150x(.5) 1  4 =57 Ibs..
=57 Ibs,
(rel.pr.)z, =57 X = 32 ! lbs  "
NOTE: In all the following problems clearance is assumed to be zero.
Prob. 1. A 12x18x24 in. engine is running on superheated steam of 150 Ibs. per
square inch absolute pressure, and exhausts to the atmosphere. If the speed is 100
R.P.M., highpressure cutoff , low pressure cutoff , and receiver volume 10
cu.ft., what horsepower will be developed and what steam used per hour?
Prob. 2. What would be the effect on the power and the economy of (a) changing i
to wet steam in the above? (6) to compressed air?
Prob. 3. What would be the receiver and the release pressures for each case?
Prob. 4. Will there be equal work distribution between the two cylinders?
Prob. 5. It is desired to obtain complete expansion in a 14x22x36in. engine
running on fluid which gives a value for s of 1.2. Initial pressure is 100 Ibs. per
square inch gage, and back pressure 5 Ibs. per square inch absolute. What must be
the cutoffs and what power will be developed at 500 ft. piston speed? Receiver =
XH.P. volume.
Prob. 6. How large must the receiver be for the above engine in order that the
pressure in it shall not fluctuate more than 5 Ibs. per sq. inch?
Prob. 7. An engine is to run on steam which will give a value of s = l.l, and to
develope 500 horsepower at 100 R.P.M. Piston speed is not to exceed 500 ft. per
minute. Steam pressure, 150 Ibs. per square inch absolute, back pressure, 5 Ibs. per
square inch absolute. Complete expansion and equal work distribution, for this load are
to be accomplished. What will be the cylinder sizes and the highpressure cutoff if the
receiver is to be 3 times the highpressure cylinder volume?
Prob. 8. What will be the steam used per hour by the engine of Prob. 7, and
what will be the variation in the receiver pressure?
Prob. 9. If the highpressure cutoff were halved, how would the power and
economy be affected?
13. Compound Engine without Receiver, Logarithmic Law. No Clear
ance, Cycle IX. General Relations between Dimensions and Work when
HighPressure Exhaust and LowPressure Admission are Coincident. Such
a peculiar case as this admits of but little modification of the cycle compared
with the receiver cases, b: cause the lowpressure expansion is necessarily a direct
continuation of the high pressure without any possible break. There can be no
overexpansion in the high nor can expansion there be incomplete, as there is,
properly speaking, no back pressure with which to compare the highpressure
cylinder terminal pressure. There may, however, be over and incomplete
expansion in the lowpressure cylinder. It might appear that the highpres
sure cylinder negative work was equal to the lowpressure admission work, as
each is represented by the area below DC, Fig. 87A, but this is not the case, since
WORK OF PISTON ENGINES
293
le diagram is drawn to two different scales of volumes, showing the pressure
jroke relation between high and low. This is apparent from the diagram, Fig.
1C showing fluid volumes in each cylinder to a single scale on which ABCD
; the work done in the highpressure cylinder, ABD'EF the whole work, whence
There is, of course,
The cycle,
)CD'EF is the part done in the lowpressure cylinder.
o lowpressure cutoff or even admission as ordinarily considered.
j far as the work to be done is concerned, is the same as for a simple engine,
H.P.Cyl. Vols.
INDIVIDUAL CYLINDER WORK SHOWN TO SAME
SCALE OF PRESSURE AND VOLUME
N K L
HIGH AND LOW PRESSURE CYLINDER DIAGRAMS PLOTTED TO SAME AXIS
'IG. 87. Work of Expansion in the Noreceiver Compound Engine, Zero Clearance, Cycle
IX, Logarithmic Expansion, Cycle X, Exponential Coincident Piston Movement.
nd the only reason for introducing formulas for overall work, work per cubic
ypt supplied, (m.e.p. referred to low), and fluid consumption, is to put them into
orm for immediate substitution of dimensional relations. Because of the absence
f cutoff in the low, the distribution of work between high and low will
lepend solely on the cylinder ratio and highpressure cutoff, for, the earlier the
aghpressure cutoff, and the larger the highpressure cylinder, the greater
lie fraction of the total work that will be done there, as there is only a fixed
mount available, and the less there will be left to be done in the low
294 ENGINEERING THERMODYNAMICS
The diagrams of the two . cylinders are plotted to combined axes in
Fig. 87 D. The points Q and R at equal heights. KN is the L. P. displace
ment, and KG that of the H.P. It has been shown in Section 8, that
the expansion lines CD and C'D' may be plotted to the axes LN and LXM,
the point X being the intersection of NQ and KR extended, and that the distance
TN RC
= ^^. DH= ^_ =Dif _l__. f . _ i (4Q2)
Hence the work area under CD is
, I J * XV "i A A / 1 \ *** H 1 */ /
GL DL DH * D H
but
hence
( 1 1
tf = 144(in.pr.)ZtfDtf lflog e ^
^H RC~
Rc\. . (404)
Again the work area under C'D' is
=^=144(rel.pr.)^ L " W e ~ t
If I *'/)_ fi & e 7^ '
hence
TFi / = 144(in.pr.)Z // D // ^^jlog e ^ c 144(bk.pr.) J D L , . (4C5)
and the total work,
W = 144(in.pr.)2W>* J 1flog, ^~ ^ loge R c
WOEK OF PISTON ENGINES
295
But
and
so that
/ Re _i\_i
\Rcl Rcl) '
loge
i **O i r>
= iOg e = = lOge HVJ
"H
. . (406)
which shows by its similarity to the work of the simple engine that, as before
stated, the total work is the same for this cycle as if the entire expansion were
made to take place in a single cylinder.
This same result could have been attained in another way sufficiently
; interesting to warrant setting it down. Since the lowpressure work is repre
sented truly to scale by C'D'EF, Fig. 87C, the mean effective pressure of the
lowpressure cylinder is given by the area divided by V e . By contracting all
'volumes proportionately, C'D' takes the position CD' and C'F the position
CF f , hence
area CD'EF
V.Vf.
P.
represents the mean effective pressure in the lowpressure cylinder just as
truly. Therefore,
" ,.' , iare&CD'EF D \ Tr
L.P. cylinder work= ^ ^ P* V e
As the highpressure work is (total low),
H.P. cylinder work = P 6 F/l+log e ^ \ P e V e .
. / f ;
r c ^ +P e F<
F
V e
Introducing symbols
L.P. cylinder
H.P. cylinder
r.)D I ,. (407)
^^ 1 \og 6 Rc\
^loge^c. . (408)
iich check with Eqs. (404) and (405),
296
ENGINEERING THERMODYNAMICS
Dividing the total work by the lowpressure cylinder volume and the high
pressure admission volume in turn,
(m.e.p. ref. to L.P.) = (in.prO
?  (bk.pr.)
(a)
 (bk.pr.) (6)
. . (409)
Re
Work per cu.ft. supplied = 144(in.pr.)Z#( l+log e ~\  (bk.pr .)R C (a)
. (410J
Cu.ft. supplied per hr. per I.H.P. = ^^f^ LJ > ^ (a)
13,750 1_
(m.e.p. ref. to L.P.) R H Rc
(&)
. . (411
For equal division of work there can obviously be only one setting of th
highpressure cutoff for a given cylinder ratio and any change of load to be me
by a change of initial pressure or of highpressure cutoff will necessarily unbalanc
the work. Equating the highpressure and lowpressure work expressions
Eqs. (404) and (405),
or
p  =
Kc~ 1
(bk.pr.)
Re
= ^  =
Kc 1
Rc+l
/bk.prA R c
( r  3?
\in.pr. / L H
Another relation exists between Z H and R c , namely, that
= Rv
f7 ~ r> >
AH 'Me
where Rv is the ratio of volumetric expansion. Then
but
hence
WORK OF PISTON ENGINES
297
With this formula it is possible to find the necessary ratio of cylinder
displacements for given initial and back pressures and for given ratio of
expansion R v .
For convenience in solving this, a curve is given in Fig. 88 to find value
2R
c
of Re when Rc R c ~ l has been found.
37
?
3
^
x^
"*^
x^
^
.
x^
'
x
^
. ^
/
/
r
X
,
/
/
, 25 50 75 100 2Rc 125 150
"Values of (Re) "*&
i. 88. Curve to Show Relation between Values of RE and (RE) R C l for Use in Solving
Eq. (412), Giving Cylinder Ratio in Terms of Ratio of Expansion for the Noreceiver
Compound Engine without Clearance.
The complete expansion case of this cycle results from the condition
P d =Pe or (r6l.pr.)ji=(bk.pr.) or R v
rhich when applied to Fig. 87, transforms the diagrams to the form Fig. 89.
It also follows that
(bk.pr.)z, = (m.pT.)Z H D a
and
Rc_ /in.pr.
~Z~ H ~ \bk.pr.
These conditions will, of course, reduce the total work Eq. (406) to the
common value for all cycles with logarithmic expansion and likewise those
for mean effective pressure, work per cubic foot supplied, and consumption.
For the equal division of work under this condition, Eq. (412), becomes
(413)
n.pr.
= i and R may represent ratio of expansion or ratio of
298
ENGINEEEING THERMODYNAMICS
initial to back pressures, these being equal. Fig. 90 gives a curve showing the
relation between cylinder ratio and ratio of expansion established by the above
condition.
H J>. Cyl. vols.
54321 V
L.P. Cyl. Vols.
\
\
FIG. 89. Special Case of Cycles IX and X. Complete Expansion in both Cylinders of the
NoReceiver Compound Engine, Zero Clearnace.
Example 1. Method of calculating Diagram, Fig. 87.
A. As described in the text this diagram is drawn to twovolume scales, so tha
there may be two volumes for one point.
WORK OF PISTON ENGINES
299
Assumed data:
p a =p b = 120 Ibs. per sq.in. abs.
p e =p f = 10 Ibs. per sq.in. abs.
To locate point C:
To locate point D:
V a = V c = V d = V g = V f =0 cu.ft.
F& = 1 cu.ft.
F c =2cu.ft.
p 120X1 A01U
PC =r 7 =  =60 Ibs. per sq.in.
r c &
p PcV c 60X2
Pa = ;; =  =24 Ibs. per sq.in.
V d O
To locate intermediate points from C to D. The volume at any intermediate
tint is (the volume of lowpressure cylinder up to that point) +( volume of high
Values of Re
M CO O1 <] <O
*
.
_^
.
*"
^
""
^
""
^
**
X"
/
/
/
/\
5 10 15 20
Values of R
EG. 90. Curve to Show Relation between Values of RC, the Cylinder Ratio, and Rihe Ratio
of Initial to Back Pressure for Complete Expansion in the Noreceiver Compound
Engine without Clearance (Eq. (413).)
pressure cylinder from that point to end of stroke), e.g., at I stroke the volume in
low, is .75x5, and the volume in the high is .25x2, or total 4.25, and the pressure
at that point is found by the PV relation as above.
B. Assumed data:
P a =p b =120 Ibs. per sq.in. abs.
P e =Pj= 10 Ibs. per sq.in. abs.
F a = F/=0 cu.ft.
F d = F e =5cu.ft.
F 6 = l cu.ft.
To locate point D:
_ P 6 F & 120X1 . .,
P d 2 =24 Ibs. per sq.in.
Va 5
Intermediate points from B to D found by assuming volumes and computing
>ressures from the PV relation as above.
C. Figure ABCD constructed as in A. Figure C'D'EF is figure CDEF of A to the
ame pressure scale but to a volume scale 2.5 times as large.
D. Figures constructed as in C.
300 ENGINEERING THERMODYNAMICS
To draw indicator cards. The volume and pressure scales are chosen and from
diagram A, a distance AB is laid off to the volume scale, AD is then laid off equal to
AD of diagram A to the pressure scale. Point C is located to these scales and joined
to B and D by drawing curves through the intermediate points plotted from the PV
diagram to the scales of the card. For the lowpressure card EF is laid off to the
volume scale, and FC' and ED' to pressure scale. C' and D' are then joined in same
manner as C and D for highpressure card.
Example 2. Find (a) the horsepower, and (6) steam used per hour for a 12 X 18 X 24
in. engine with no clearance when initial pressure is 150 Ibs. per square inch absolute,
back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cutoff in
the highpressure cylinder is , there being no receiver.
(o) From Eq. (409) we have
1
(m.e.p.)=(in.pr.) p p [1 +log e (###c)] (bk.pr.),
10=50 Ibs. sq.in.
hence
I.H.P.=192.
(b) From Eq. (411) we have
13 750 1
Cubic feet of steam per I.H.P. per hour =7 X ^,
(m.e.p.) R H ttc
_13,750 _^_
50 X 2X2.25
hence the weight of steam used per hour will be
61 .2 X .332 X 192 = 3890 pounds.
Example 3. What will be the cylinder ratio and the highpressure cutoff to give
equal work distribution for a ratio of expansion of 6, an initial pressure of 150 Ibs. per
square inch absolute and back pressure of 10 Ibs. per square inch absolute?
Ratio of back to initial pressures is .067 and
hence from Eq. (412)
2R C
loge 7 ^ 1.40,
or
2R C
Rc R c i =24.36,
and from Fig. 88
7? *? &
From the relation Z# = c = highpressure cut off = r = .446.
liv o
WORK OF PISTON ENGINES 301
Prob. 1. A compound locomotive has no receiver and runs on an initial pressure of
175 Ibs. per square inch gage and atmospheric exhaust. The cylinders are 18 and
50 X42 in. The steam pressure may be varied, as may also the cutoff to a limited
legree. For a speed of 200 R.P.M. and a constant cutoff of f , find how the power
ivill vary with initial pressure and for constant initial pressure equal to boiler pres
sure show how the power at the same speed will vary from % cutoff to full stroke.
Prob. 2. Show how the steam used per horsepower hour will vary in above
problem.
NOTE: d for 175 Ibs. =.416.
Prob. 3. With the cutoff at f, what should the initial pressure be to give equal
work distribution?
Prob. 4. With full boiler pressure and J cutoff what would be terminal pressure
in the lowpressure cylinder?
Prob. 5. What must be size of cylinders for a tandem compound engine with
negligable receiver volume to run at 125 R.P.M. with complete expansion and equal
work distribution on an initial pressure of 125 Ibs. per square inch gage and a back
pressure of 5 Ibs. per square inch absolute, when carrying a load of 500 horsepower,
'the piston speed to be less than 500 ft. per minute?
Prob. 6. What will be the steam used by the above engine in pounds per hour?
NOTE: 8 for 125 Ibs. =.311.
Prob. 7. A builder gives following data for a tandem compound steam engine.
Check the horsepower and see if the work is equally divided at the rated load.
Cylinders 10 ins. and 17^X15 ins., initial pressure 125 Ibs., speed 250 R.P.M., horse
power 155. Neglect the receiver volume.
Prob. 8. Another manufacturer gives for his engine the following, check this:
Cylinders 20 and 32x18 ins., initial pressure 100 Ibs., atmospheric exhaust, speed
200 R.P.M., horsepower 400. Neglect the receiver volume.
14. Compound Engine without Receiver, Exponential Law, Cycle X.
General Relations between Dimensions and Work when Highpressure
Exhaust and Lowpressure Admission are Coincident. Referring to Fig.
87 D it is desirable first to evaluate the work areas CDKG and C'D'NK. As
before,
t
= D H ~r and
hence
(rel.pr.)#=(in.pr.)Zj/,
?[!_ /JV 1
l \R C )
302 ENGINEERING THERMODYNAMICS
whence
..  .r.)^. . . (415)
81 L Kc 1 J
It is to be expected that the sum of high and lowpressure work will be
of a form similar to that which would be obtained if all work were performed
in a single cylinder of a displacement equal to that of the low pressure, adding,
W = WH+W L = 144(in.pr.)
DHZ H \ _ 7 ,.,
LI L V Rci
144(bk.pr.)Dz,
r /ivni
+ZJST l  ^j II 144(bk.pr.)I>z,
7 1
whence, substituthig D^=IT
r )^; (416)
f rft TTM (m.pr.) Z g f /M 8 " 1 !
: ^^~5;L S (^J J(bk.pr.) ..... (417)
Work per cu.ft. supplied = 144 { m *" ') L_ (i) 8 " 1 144(bk.pr.) (418)
L \F/ J Z ff
Cu.ft. supplied per hr. per I.H.P. = 13 ' 750 ??
(m.e.p. ref. to L.P.) R c
13,750
(mie.p. ref. to L.Pj R v ' ' ^ 41
Conditions for equal division of work between high and lowpressure
cylinders may be obtained by equating Eqs. (414) and (415).
sZ H *Z H i
= ZiR c \ A^_l t**T.\R c
L Rc1 J WprJ^ 5 " 1 )*
WORK OF PISTON ENGINES 303
Rearranging
The last term in the first member of this equation may be expressed as
/bk.prA,, ,
( }Rv(s 1)
\in.pr. )
and the relation
Re
IH=
xists between Z H and Re, hence, making these substitutions,
^'W(DI (420)
which is not a simple relation, but can be solved by trial.
The assumption of complete expansion in the lowpressure cylinder (it is
(ways complete in high, for this cycle), leads to this following relations:
/in.pr. \ _ f
I in I K>v t
\bk.pr./
I hence
144(bk.pr.)D L = 144(in.pr.)^D^,
and from Eq. (414),
TF*=144(in r )^ \s ( V' 1  ']
o L \ ' / " * 1
but
R c Rysl s1 1 /bk.pr.Xs
p=Rv, and ^ ^p~r = ^ s v and ' ^~ = \in~V/
sl
... (421)
(m .e.p. ref. to L.PO^in.pr.^l. . . (422)
304
ENGINEERING THERMODYNAMICS
The expression for equality of work Eq. (420) becomes, for this case of complete
expansion,
Rcl
Rcl
. . . (423)
by which it is not difficult to find the ratio of expansion R v , which gives equality
of work for given values of s, and Re, the cylinder ratio. Values for R v for
various values of R c and s are given by the curves of Fig. 91.
10
*
s
I 5
1.5
1.4
Values of b
1.3 12
1.1
10
Values of Rv
15
FIG. 91. Curves to Show Relation between R c the Cylinder Ratio, and H v the Ratio of
Expansion, for Various Values of (s), Applied to the Noreceiver Compound Engine
without Clearance, when the Expansion is not Logarithmic.
Example 1. Find (a) the horsepower, and (6) the steam used per hour for a
12 and 18x24in. engine with no receiver when the initial pressure is 150 Ibs. per
square inch absolute, back pressure 10 Ibs. per square inch absolute, speed 125
R.P.M., cutoff in highpressure cylinder is J, there being no receiver and steam
having expansion, such that s = 1.3.
From Eq. (417)
which, on substituting the above values, becomes
sq.in.
hence the indicated horsepower =243.
WORK OF PISTON ENGINES 305
(6) From Eq. (419) the steam used per hour in cu.ft. per horsepower is
.
'
13,7502^
m.e.p. Re
which, for the data given above, becomes
13,750 .5
or pounds per hour total, is, 48.2 X 243 X. 332 =3880.
Example 2. What will be the highpressure cutoff and cylinder ratio to give
equal work distribution and complete expansion for an initial pressure of 150 Ibs. per
square inch absolute, and back pressure of 10 Ibs. per square inch absolute?
From relation # F s = (r^ ), #F = 6.9 and from this, by the curve of Fig. 91,
\DK.pr./.
For complete expansion
Re 5.4 _
~~~
Prob. 1. A tandem compound engine without receiver has cylinders 18 and
X42ins. and runs at 200 R.P.M. What will be the horsepower developed at
ds speed if the initial pressure is 175 Ibs. per square inch gage, back pressure
itmosphere, highpressure cutoff ^, and s has a value of (a) 1.1, (6) 1.3? Compare
the results with Prob. 1 of Sec. 13.
Prob. 2. What will be the weight of steam used per horsepower per hour for
two cases of the above problem? Compare these results with those of Prob. 2,
>c. 13.
NOTE: B = .416.
Prob. 3. What must be the cutoff in a 10 and 15 X20in. compressed air engine
running on 100 Ibs. per square inch gage initial pressure and atmospheric back pres
sure, to give complete expansion, and what will be the horsepower per 100 ft. per
minute piston speed, s being 1.4?
Prob. 4. It is desired to run the following engine at its most economical load.
What will this load be and how much steam will be needed per hour?
Cylinders 20 and 32x18 ins., speed 150 R.P.M., steam pressure 100 per square
inch gage, atmospheric exhaust, dry saturated steam.
Prob. 5. Should the load increase 50 per cent in Prob. 4, how would the cutoff
change and what would be th effect on the amount of steam used?
Prob. 6. What would be gain in power and the economy of the engine of Prob. 4
were superheated steam used, for which s = 1.3?
Prob. 7. In a 14 and 20x24in. engine will the work be equally divided
between the cylinders for the following conditions? If not, what per cent will be done
in each? Steam pressure 100 Ibs. per square inch absolute, back pressure 10 Ibs. per
square inch absolute, s = 1.2, cutoff = 5.
Prob. 8. What would be the work and steam used by the above engine if there
were complete expansion and equal distribution?
306
ENGINEERING THERMODYNAMICS
15. Compound Engine with Infinite Receiver. Logarithmic Law. With
Clearance and Compression, Cycle XI. General Relations between Pressures,
Dimensions and Work. In terms of point pressures and volumes, Fig. 92, the
INDICATOR CARDS OF EQUAL BASE
AND HEIGHT FOR CASE OF INCOM
PLETE EXPANSION AND COMPRESSION.
INCOMPLETE EXPANSION ANCT COMPRESSION. V
 ( rec.pr ) R
(in.pr.
*4 rel.pr
bk.pr.)
( rel.pr.)
INDICATOR CARDS OF EQUAL BASE
AND HEIGHT FOR CASE OF OVER
EXPANSION AND COMPRESSION.
OVER EXPANSION AND COMPRESSION.
FIG. 92. Work of Expansive Fluid in Compound Engines with Infinite Receiver, with Clearance,
Cycle X Logarithmic, and Cycle IX Exponential Expansion, and Compression.
work of the two cylinders may be written down at once as if each were inde
pendent of the other, the connection between them being fixed first by making
the back pressure of the high equal to the initial pressure of the low, or to
WORK OF PISTON ENGINES
307
he receiver pressure, and second by making the volume admitted to the low
qual to that discharged from the high reduced to the same pressure. This
ist condition may be introduced in either of two ways,
(a) EM = NH,
(b) [(PV) on H.P. expansion line (P V) on H.P. comp. line],
= [(PF) on L.P. expansion line (PV) on L.P. comp. line].
Without introducing the last relation
^^^ (424)
^)PzF;(^ . (425)
+P, t V h P l V l P V +P l V l PjV J +P k V k .
The second condition is
or
Substituting
10g e ~
Vb
log, ~PcVe 10g e ^~P*V t log, ^
Vh * f
(426)
+2(P b V b +P k V k )P a V a P g V g P d V d PjV J
This expression, Eq. (427) contains, however, the receiver pressure which is
ited to the release pressure by
(rec.prO =P^Pe = P^P. = fi=
introducing this
W =
, log e ~ P *
(428)
308 ENGINEERING THEEMODYNAMICS
Introducing the usual symbols in Eqs. (427) and (428) and in additic
the following:
Z = cutoff as fraction of stroke, so that Z H D H is the displacement volun
up to cutoff.
c = clearance volume divided by displacement, so that C H D H is the clearan<
volume and (Z H +c H )D H is the volume in the highpressure cylindi
at cutoff.
X = that fraction of the stroke during which compression is taking place i
that (X H +Cn)D H is the volume in the highpressure cylinder whe
compression begins.
Applying the general symbols to Eq. (427),
(in.pr.) (Z H +c H )D H log* (J~^)
+ (Tec.pr.)(Z L +c L )D L log e
W = 144
 (rec.pr.) (X H +c ff )D H log*
 (bk.pr.) (
+2(in.pr.) (Z H +c a )D H  (m.pr.)c H D H  (rec.pr.)
+2(bk.pr.) (X L +c L )D L  (rec.pr.)c L Z>  (bk.pr.) (1 +c L }D L
. (4!
This expression gives the work in terms of initial, receiver and back pressur
the valve periods, cutoff and compression, the clearances and cylinder c
placements.
Substitution of the symbols in Eq. (428) will give another equivalent expr
sion in terms of the same quantities except that lowpressure cylinder reles
pressure will take the place of receiver pressure. This is
/ , log*
, /l+c L \ /X H +D H \
 (rel.pr.)/, ,. (X H +c H )D H \og e I 
\/L+CZ,/ \ C H )
 (bk.pr.) (X L +c L )D L log,
+2(in.pr.)(Zjsf +c H )D H  (in.pr.)c^D^  (pel.pr.)z.( J + L  )
\ZL +CL I
WORK OF PISTON ENGINES 309
'' It is sometimes more convenient to involve the cylinder ratio and low
pssure displacement than the two displacements as involved in Eq. (430)
id the ratios of expansion instead of cutoffs. This may be done by
iJ T
V c
d it should be noted here that the ratio of expansion in each cylinder is no
iger the reciprocal of its cutoff, as was the case when clearance was zero, nor
the whole ratio of expansion equal to the product of the two separate ones
cause the lowpressure cylinder expansion line is not a continuation of that
the high. Making these substitutions for cylinder and expansion ratios,
[. (430) becomes,
log e Rn + (rel.pr.)z,(l +C L ) log e R L
44D/,
^(l +C H )
+2(bk.pr.) (X L +C L )  (relpr.) L R L c L  (bk.pr.
It is interesting to note that this reduces to Eq. 304 of Section 9, by making
ance and compression zero.
From any of the expressions for work, but more particularly (430) and
52), the usual expressions for (m.e.p.) referred to lowpressure cylinder, work
3r cubic foot supplied, and consumption per hour per I.H.P. can be found,
it as these are long they are not set down, but merely indicated as follows:
W
(m.e.p. ref. to L.P.) = . ..   . . (433)
W
fork per cu.ft. supplied =
u.ft. sup. per hr. per I.H.P.
13,750 , rec.
~ 
13,750 . , Y , f )
~ (XL+CL) rec.pr . n.pr.
310 ENGINEERING THERMODYNAMICS
As the receiver pressure is related to the initial and back pressures and t<
the relation between the amount taken out of the receiver to that put in, whicl
is a function of the compression as well as the cutoff and cylinder ratio, it i
expressed only by a complicated function which may be derived from th(
equivalence of volumes in the high and low, reduced to equal pressure.
PVP e V e = P h V h ^P t V t , or P ft
Therefore,
Tvftv
Introducing symbols
c.pr.) = (in.pr.) (ZL+CL )^+^^C^D^ + (bk '^ (Z^c L )D L +(X H +c H )D a
Hence
4 (bk nr ^ (^L+c L )R c ,.
h^DK.pr.;, \v__i_fY i r^ ^ 4t5b
This Eq. (436) gives the receiver pressure in terms of initial and back pressures
the two clearances and compressions, the cylinder ratio and the cutoff in eac
cylinder.
Proceeding in a similar way, the release pressures can be found in terms
initial data,
17
p P c
t C * &T7 J
v b
or
(a)
(43
(6) '
And
or
(Xt+c L )
l.pr.), = (m.pr.)^^^^ +(bk.pr.)
(43!
WOKK OF PISTON ENGINES
311
These three pressures all reduce to those of Eqs. (308), (309), (310), Section
9, when clearance and compression are zero.
Equal work in both cylinders is, of course, possible, but it may be secured
by an almost infinite variety of combinations of clearance, compression and
cutoff in the two cylinders for various ratios of expansion; it is, therefore, not
worth while setting down the equation of condition to be satisfied, but reference
may be had to Eqs. (424) and (425), which must be made equal to each other, the
result of which must be combined with the equation of cylinder relations.
\
\
INDICATOR CARDS OF EQUAL
BASE AND HEIGHT
FIG. 93. Special Case of Cycles XI and XII Complete Expansion and Compression in both
Cylinders, of Compound Engine with Clearance and Infinite Receiver.
There are certain special cases of this cycle for which equations expressing
important relations are simpler, and they are for that reason worth investigat
ing. Those that will be examined are
(a) Complete expansion and compression in both cylinders, Fig. 93.
(&" Complete expansion in both cylinders with no compression, any clearance,
Fig. 94.
(c) Any amount of expansion and compression but equal in both cylinders,
equal clearance percentages and a cylinder ratio equal to the square root of
the ratio of initial to back pressures, Fig. 95.
Case (a) When both expansion and compression are complete in both cylin
ders, Fig. 93,
.pr.)^*log. , .... (439)
A, .... (440)
312 ENGINEERING THERMODYNAMICS
but
Z#D//(in.pr) =Z L D L (rec.pr.)
and
(in.pr .)\ /(rec.pr.)\ /(in.pr.) (rec.pr.)\ _ (in.prO
p7)/ H  logc l(bk^)/~ 10gc l(r^rO (bk.pr.)/ ge (bk.pr.)'
hence
'"irriV '  : ' (441)'
(442)
Work per eu.ft. supplied = 144(in.pr.) log,  ....... (443)
rr
Consumption, cu.ft. per hr. per I.H.P^^ l__ 5. . . . (444)
(m.e.p. ref. to L.P.) R c
Equality of work in high and lowpressure cylinders is obtained by making
/ (in.pr. )\ = /(recprOx /(in^j_)\ *
Urec.pr.)/ \(bk.pr.)/ V(bk.pr.)/ '
or
(rec.pr.) = [(in.pr .)(bk.pr.)]* ....... (445)
It is desirable to know what clearances and displacements will permit of
equal work and complete expansion and compression.
hence
or calling
(in.pr. )
/(JnprOx = ^CH_ = /(in.pr. )*\
\(rec.pr.)/ Z H +c a \(bk.pr.) /'
/(rec.pr.)\ = l+c L = /(in.pr. )\ *
\(bk.pr.)/ Z L +c L \(bk.pr.)/ :
Equatuig discharge of high and intake of lowpressure cylinders,
or = R C =R.
WORK OF PISTON ENGINES
Inserting in this the values just found for Z H and Z L ,
D ,
313
(447)
rtiich is the required relation between cylinder sizes, clearances and ratio of
>ressures, which, together with cutoffs given in Eq. (446), will give equal work
nd complete expansion and compression. The compression in the high
ressure cylinder is such that
nd for L.P. cylinder.
P
(448)
L A
\
\
H\
G H
INDICATOR CARDS OF EQUAL
BASE AND HEIGHT
V
lop. 94. Special Case of Cycles XI and XII. Complete Expansion and Zero Compression
in both Cylinders of Compounds Engine with Clearance and Infinite Receiver.
I Case (b) With complete expansion and no compression, both cylinders, any
jearance, Fig. 94,
1 (449)
T^'T) c [(rec.pr.)(bk.pr.)]l (450)
^DK.pr. ;/ j
T
^DK.pr.
8th the added requirement_that the highpressure discharge volume, EC = low
essure admission volume FH, or
(451)
314 ENGINEERING THERMODYNAMICS
and
...... (452)
hence
[(bk.pr. )] = Zz,+Cz,
L(rec.pr.)J l+c L '
which substituted in Eq. (451) and rearranging gives
R c = 
Eq. (453) indicates that for this special case of complete expansion and no com
pression the cylinder ratio required to give this case, is determined entirely bj
the L.P. cutoff and clearance. If the cylinder ratio and clearance are fixed, th(j
required cutoff in the L.P. cylinder can be found by solving Eq. (453) for Z 2
Z L =^f^c L , (454
tic
and from Eq. (452),
(rec.pr.) = (bk.pr.)
D
ti
Cutoff in the highpressure cylinder is determined by clearance, initia
pressure and receiver, pressure, which in turn depends on lowpressure cutoi
and clearance Eq.(452), or may be reduced to cylinder ratio and lowpressur
clearance by Eq.(454), as follows
V c= I+CH = /( in.pr. )\ = /(in.pr. )\Z L +c L
V* Z H +CH \(rec.pr.)/ \ (bk.pr.)/ l+c L '
hence
,, (l+c s )(l+c L )
AH= p (7  x  CH
Rp(^Lrc L )
Eliminate Z L by Eq. (454),
Re ,
" cir  (45(
Since the high and lowpressure cutoffs are functions of cylinder and clearam
dimensions, and of Rp, the rato 'of initial and back pressures, the work of higl
and lowpressure cylinders may be expressed entirely in terms of these quantities
(45!
WOKK OF PISTON ENGINES
Hence, total work by addition is
315
c L (R c l)\. . . (459)
Expressions might be easily written for mean effective pressure referred
to the L.P. cylinder, work per cubic foot fluid supplied, and consumption, but
will be omitted for brevity. It is important to note, however, the volume of
fluid used per cycle is not AB, but is LB, Fig. 94, and is,
(Sup.Vol.) = D H \ (Z H +c H )  C H ^ ^1 = D H [ (Z H +c H )  cj^ I . . (460)
L (m.pr.; j L Mr A
(m.e.p.ref.toL.P.)= ni ^ c ....... (461)
w
(Work per cu.ft. supplied) = ( gup y ol y .... (462)
13 750 f
Consumption cu.ft per hr. per I.H.P. = / m r ^ f to L P ) &*
Equality of work, secured by equating Eqs. (457) and (458) gives
cc.^el).  (464)
This equation may be satisfied in an infinite number of ways. One case
worth noting is that of equal clearances, when it is evident that if
= CL, and = #c, or
the Eq. (464) is satisfied. This last condition is the same as that which satisfied
Case (a) with complete compression.
Case (c), Fig. 95, assumes that
and
316
ENGINEERING THERMODYNAMICS
and corresponds to the first special case considered in Section 9, which lead
in the noclearance case to equality of high and lowpressure work.
FIG. 95. Special Case of Cycles XI and XII, Equal Per Cent Clearance in Each Cylinder of
Compound Engine with Infinite Receiver and Cylinder Ratio Equal to the Square Root
of Initial Divided by Back Pressure.
The assumptions already made are sufficient to determine the receiver pres
sure. By Eq. (436)
(rec.pr.) = (in.pr.)
Z+c
n.
+ (bk.pr.)
[(in.pr.) (bk.pr.)]*
= [(in.pr.) (bk.pr.)]*. . . . ....... . . . . .
The work of the highpressure cylinder may now be evaluated.
^=144D // (m. P r.){z[nlo g e (^)1~
{ L w~rC/j
144D^[(in.pr.) (bk.pr .)]*
. (465)
. (466)
WORK OF PISTON ENGINES
The lowpressure cylinder work may be similarly stated,
but
~D*l=^
317
(467)
 144D (bk.pr.) I (X+c) log e ( ~^) +1 X \ ,
\ c /
and similarly,
.) =D*[(in.prO(bk.pr.)] x .
With these substitutions the value of lowpressure work,W L ,lq. (467), becomes
equal to high pressure work, Eq. (466), hence the total work
(468)
Example 1. Method of calculating Diagrams, Fig. 92.
Assumed data:
P q =P a =Pb = 120 Ibs. per square inch abs.
p n =p g =p e =p d =p h = 50 Ibs. per square inch abs.
P t =Pj= 10 Ibs. per square inch abs.
(V h V n )=(VmVe).
F=y/=.12cu.ft.
F 6 = .4cu.ft.
V c = V d = .Scuft.
V g = Fi = .16cu.ft.
F <= V= 2cu.ft.
V k = Acu.it.
The above may be expressed in initial pressure, etc., and in terms of cutoff, etc.,
but as the relation of the lettered points to these terms is shown on the diagram values
for cutoff, etc., they will not be given here, as they may readily be found from values of
the lettered points.
To locate point C:
To locate point F:
To locate point Q:
=60 Ibs. per sq.in.
V f .12
PeF e= ^0 =
* Q rk * r\s\
per sq.in
318 ENGINEERING THERMODYNAMICS
To locate point L:
 T7 1C =25 Ibs. persq.m.
Vi .lo
To locate point N:
F n = = = . 08 cu.ft.
To locate point 77:
(7 A _F B ) = (7 OT _F e ), or 7 / , = F m + 7 n F e = .96+.08.2 = .84 cu.ft.,
since
PmV m = P b V b , = V m ~ = .96 CU.ft.
To locate point 7:
Example 2. Find (a) the horsepower, (6) steam used per hour, and (c) receiver
and release pressures for a 12 and 18x24in. engine with infinite receiver, 6 per
cent clearance in highpressure cylinder, and 4 per cent clearance in lowpressure cylinder,
when initial pressure is 150 Ibs. per square inch absolute > back pressure 10 Ibs. per
square inch absolute, speed 125 R.P.M., cutoff in highpressure cylinder is f, low
pressure cutoff is such as to give complete H.P. expansion, and compression is 15 per
cent in high and complete in low.
(a) For complete highpressure expansion the receiver pressure must be equal to the
highpressure release, and to maintain the receiver pressure constant the lowpressure
cylinder must take as iaich steam per stroke as the highpressure discharges. With
initial pressure and cutoff as given, the release pre:sure for the highpressure cylinder
may be found from the relation (in pr.)(c#+Z#) =(re\.pT.) a (c H \D H ) or 150x(.56)
= (rel.pr.)//(1.04), or (rel.pr.)H = 79.3 Ibs. Since there is 15 per cent compression in
highpressure cylinder there is exhausted each stroke 85 per cent of its volume. Also
since compression in lowpressure cylinder is complete, the lowpressure clearance is
full of steam at the receiver pressure at the beginning of the stroke. Hence the
lowpressure displacement up to cutoff must equal .S5D H or L.P. cutoff = .85D H ,
divided by cylinder ratio, or .85 J 2.25 = .378. As compression is complete, the
per cent compression may be found from the relation Cz,x(rec.pr.) =(c L +X L )(bk.pT ),
or .04X79.3 = (.04+Xz,)10, or Xz, = .28.
From Eq. (432), (m.e.p.) referred to lowpressure cylinder is obtained by dividing
by 144 D L , and on substituting the above values it becomes,
150(.1+.06)( ) log, 2+30(1 +.04) log, 2.6430(.15 + .06) (~)log e (^
10(.28+.04)lo ge
"264
30X^(1+.06)+2X10(.28+.04)30X2.64X.0410(1+.04)=60.5 Ibs. per. sq.in.,
hence
I.H.P.=235.
WORK OF PISTON ENGINES
(6) From Eq. (435) by substituting the above values
13 750 T /
u.ft. steam per hour per horsepower = ' ( (.
o0.5 I \
319
38 +.04) (.28 +.04)  
loO
1=45.5,
O J
pounds per hour will be 3550.
(c) Release pressure for highpressure cylinder has been shown to be 79.3 Ibs.
may be checked by Eq. (437), as follows:
79.3 Ibs.
Receiver pressure has already been shown to be equal to this quantity and may
checked by Eq. (436)
(rec.pr.)//
150 X (.5 +.06)
_ IPX (.28 +.04)2.25
(.378 + .04)2.25 + (.28+.04) + (.378+.04)2.25 + (.28+.04)
= 79.3 Ibs.
Lowpressure release pressure is found from Eq. (438) to be
.28 + .04
i _i_ r\A. 9 \xO OF;
(rel.pr.)/,
1 +
(.15 + .06)2.64
_ (1 +.04)2.25 _
+ 10
1+.04
.06)2.64
1+ (1+. 04)2.25 _
= 30 Ibs.
Prob. 1. What will be the horsepower and steam used by the following engine
or the data as given?
Engine 20 and 28x36 ins., running at 100 R.P.M., clearance 5 per cent in high
>ressure, 3 per cent in low. From cards H.P. cutoff = .3, L.P. = .4, H.P. compression,
1, L.P., .2. Gages show (in.pr.) to be 150 Ibs., (r c.pr.) 60 Ibs., (bk.pr.) 26 ins.
Ig. (barometer = 30 ins.).
Prob. 2. What must be ihe cutoffs and the cylinder ratio of an engine to give
qual work and complete expansion and ompression for 200 Ibs. per square inch
.bsolute initial pressure and atmosphe: ic exhaust, if clea ance is 5 per cent in the
igh and 3 per cent in the lowpressure cylinder? What will the horsepower for an
Bgine with a lowpressure cylinder 24x36 ins., running at 100 R.P.M. for this case?
Prob. 3. Should there be no compression, how would the results of Prob. 2 be
tered?
Prob. 4. What will be the total steam used by engines of Probs. 2 and 3?
Prob. 5. For an 11 and 19x24in. engine with 5 per cent clearance in each
ylinder, cutoff in each cylinder, and 20 per cent compression in each cylinder, what
rill be the horsepower and the steam consumption when the speed is 125 R.P.M., the
nitial pressure 150 Ibs. per square inch gage, and back pressure at atmosphere?
; 16. Compound Engine with Infinite Receiver. Exponential Law, with
Clearance and Compression, Cycle XII. General Relation between Pressures,
dimensions and Work. Referring to Fig. 92, of the preceding section, which
320 ENGINEERING THERMODYNAMICS
will represent this cycle by a slight change of slope of the expansion and cor
pression lines, the highpressure work may be expressed in terms of dimensior
ratios and pressures. Since this must contain receiver pressure as a factc
and since that is not an item of original data, it is convenient first to sta
receiver pressure in terms of fundamental data:
But
V?^ and Vn
rec.pr./ \rec.pr
Whence,
whence
c.pr./ \rec.pr.
or in terms of dimensions and pressures,
i
(rec.pr.) = (in.pr.)l  R (z
The highpressure work may be stated as follows:
C H
WOEK OF PISTON ENGINES
321
. . (472)
(473)
The expression for total work need not be written here, as it is simply the sum
>f Eqs. (470) and (472) or of (471) and (473), the former containing receiver pres
sure and the latter containing only dimensions, initial and back pressures and
both, the exponent of the. expansion, s,
The volume of highpressure fluid supplied per cycle is QB, Fig. 92, which
may be expressed either in terms of high pressure or of low pressure points,
thus;
(Sup.Vol.)
(474)
The following quantities will be indicated, and may be evaluated by sub
stitution from the preceding:
W
W
W
Work per cu.ft. fluid supplied = ^ y o \ )
sumption cu.ft. per hr. per I.H.P.
13,750
(Sup.Vol.)
(m.e.p.ref.toL.P.) D L
Equal division of work between high and lowpressure cylinders requires
[that Eqs. (470) and (472), or (471) and (473) be made equal. The latter will
322 ENGINEERING THERMODYNAMICS
give an expression showing the required relation between dimensions and initial
and final pressures, cutoff and compression in high and lowpressure cylinders.
In this expression there are so many variables that an infinite number of com
binations may be made to give equality of work.
It is desirable to examine the results of assuming special conditions such as
those of the previous section, the most important of which is that of complete
expansion and compression in both cylinders, which is represented by Fig. 93.

.. . . (478)
81
' ], . , . (479)
but
i_
in  r 
hence
/rec.pr.\ /jn^pr^y [ 1 _/bk : Pi\ s 1 1
\ in.pr. / \rec.pr. / [ \rec.pr ./ J J
[m.prOD^^J^/rec^XT 1
t \in.pr. /
Sl 81 S l
rec.pr. \ s _ / recjjr.X * /bk.pr.X a
, in.pr./ \in.pr7/ \rec.prT/ J
sl
(480
The receiver pressure may be found as follows. In Fig. 93, EC = GH:
Equating
WORK OF PISTON ENGINES
323
When this is solved for receiver pressure it results in an equation of the
econd degree, which is somewhat cumbersome, and will not be stated here.
]q. (481) is, however, used later to find Re.
If work is to be equally distributed between high and lowpressurecylinders,
rom Eqs. (478) and (479),
Sl 81 81 Sl
_ /rec.prA _ /rec.prA * _ /rec.prA /bk.pr^N *
\ in.pr. / ~ \ in.pr. / \ in.pr. / \rec^prV '
Sl Sl
rec.prA . =1 /bk.pr.\ .
in.pr. / \in.pr. / '
1.0
25
50
tr i f
Values o
75
100
}. 96. Curves to Show Receiver Pressure to Give Equal Work Distribution when Expansion
and Compression are Complete in both Cylinders of the Compound Engine with Infinite
Receiver, with Clearance when Expansion and Compression are not Logarithmic.
ice, for equal division of work,
(rec.pr.) = (in.pr.)
/bkpr.\Vl^i
\in.pr./
(482)
lich, if satisfied, will give equality of work in the two cylinders, for this case
perfect compression and expansion.
In Fig. 96, is given a set of curves for use in determining the value of the
>io of (rec.pr.) to (in.pr.) as expressed by Eq. (482).
324 ENGINEERING THERMODYNAMICS
When (rec.pr.) has been found by Eq. (482) it is possible by means of (481)
and the clearances to find Re The events of the stroke must have the follow
ing values to maintain complete and perfect compression and expansion.
: (483)
(484)
(486)
Example. Find (a) the horsepower, (b) compressed air used per hour, and (c) receiver
and relea.e pressures for a 12 and 18x24in. engine with infinite receiver, 6 per centi
clearance in the highpressure cylinder, and 4 per cent in the lowpressure cylinder, when
initial pressure is 150 Ibs. per square inch absolute, back pressure 10 Ibs. per square
inch absolute, speed 125 R.P.M., cutoff in highpressure cylinder , lowpressurei
cutoff such as to give complete expansion in highpressure cylinder, compression inj
highpressure cylinder 15 per cent, and complete in low. Expansion such that s = 1.4.;
(6) As in example of Seztion 15, receiver pressure equal highpressure release
pressure, and lowpressure volume at cutoff must equal volume of steam exhaustec
from high pressure. Release pressure may be found from relation (m.pY.J(cH+Z H )
= (rel.pr.) (c#+Z)//) s . or 150(.06+.5) 14 = (rel.pr.) (.06+1) 1 ' 4 , or(rel.pr.) =60 Ibs. A
in the previous example, the lowpressure cutoff is .38, and the lowpressure compressior
may be found from the relation c/, s (rec.pr.) = (cz,+Xz,) s (bk.pr.),or(.04) 14 x60
(.04+Xz,) 1 ' 4 (10), or X L = .09.
From the sum of equations (471) and (473) divided by U4D H Rc, and with th
proper values substituted the following expression for (m.e.p.) results;
I/ 10 \ .71 \ i.
.5 + .06 + 2.25(.15 + .06)()
2.25(.38+.04)+.15 + .06
.71 1 1.
^ (^)'N+*M :
.4 L\ .06 / ( 2.25(.38 + .i
f.38+.04r /.38 + .04V 4 ]
L'ni+wj . <
.04
~ar  1 + 1  09 =55 Ibs. persq.in.
hence the horsepower is 214.
WORK OF PISTON ENGINES 325
From Eq. (477) with proper values substitute,
Cu.ft.perI.H.P. hr.^pxTu
oo (_
total steam per hour will be
50X214 = 10700 cu.ft.
(c) Release for the highpressure cylinder has already been given as 60 Ibs. and the
ceiver pressure the same. The latter quantity may be checked by equation (469)
id will be found to be the same. The lowpressure release pressure may be found
jm the relation (rec.pr.)(Z i +cz / ) 1  4 = (rel.pr.) z/ (l+c z ,) 1  4 , which on proper substitution
ves
.38+.04\fi
(rel.pr.)z,=60( r ) =27 Ibs. per sq. inch
\ 1 .U4 /
 Prob. 1. What will be the horsepower and steam used per hour by an 18 and
X30in. engine with 5 per cent clearance in each cylinder and with infinite receiver
inning on 100 Ibs. per square inch gage initial pressure, and 5 Ibs. per square inch
:>solute back pressure, when the speed is 100 R.P.M. and the cutoff in highpressure
linder is \ and in low T 4 W ?
NOTE: s = 1.3 and 8 = .2.
Prob. 2. What must be the receiver pressure for equal work distribution when the
iitial pressure has the following values for a fixed back pressure of 10 Ibs. per
tuare inch absolute? 200, 175, 150, 125, 100, and 75 Ibs. per square inch gage?
Prob. 3. For the case of 150 Ibs. per square inch gage initial pressure and 14 Ibs.
If square inch absolute back pressure, what will be the required highpressure cylinder
peforan air engine with a lowpressure cylinder 18x24 ins., to give equality of work,
jjarance in both cylinders being 5 per cent?
Prob. 4. What will be the horsepower and air consumption of the above engine
len running at a speed of 150 R.P.M. , and under the conditions of perfect expansion
d compression?
, 17. Compound Engine with finite Receiver. Logarithmic Law, with
I'earance and Compression, Cycle XIII. General Relations between Pressures,
mensions, and Work when H.P. Exhaust and L.P. Admission are Inde
indent. As this cycle, Fig. 97, is made up of expansion and compression lines
jferred to the different origins together with constant pressure, and constant
lume lines, the work for high and lowpressure cylinders and for the cycle can
j set down at once. These should be combined, however, with the relation
S'ted for the case of infinite clearance which might be termed the condition
T a steady state
[(PV) on H.P. expansion line][(PF) on H.P compression line]
= [(PV) on L.P. expansion line PV on L.P. compression line,]
P h V b PeV e = P fl V fl P t V t (487)
326
ENGINEERING THERMODYNAMICS
INCOMPLETE EXPANSION ANC
(in.pr.) L
(c.o.pr.)L
I (rel.pr.) L
l(rel.pr.) u <
OVER EXPANSION AND COMPRESSION
FIG. 97. Work of Expansive Fluid in Compound Engine with Finite Receiver and w
Clearance. Cycle XIII, Logarithmic Cycle XIV, Exponential Expansion, and Co
pression.
WORK OF PISTON ENGINES 327
Besides this there is a relation between H.P. exhaust and L.P. admission
pressures, corresponding to the. equality that existed for the infinite receiver,
ithat may be set down as follows:
P m V m = PV, and P m (V m +0)=P d (V d +0)',
.'. P m V m = P b V b = P d (V d +0)P m O, and P m = P h ',
p b v b +p h o
*~ v d +o
Also
( }
, and P n (V n +0)=P g (V g +0);
P n O, and P n = P e ,
V a +0
These two expressions for the pressure at D and at G are not available in
leir present form, since they involve two unknown pressures those at H and
, but two other equations of relation can be set down from which four equa
jtions, the four unknown pressures P e , P d) P g and Pn, can be found. These
mother equations are
Pt(V d +0)=P e (V e +0), or p r P.(Mg), . . (490)
;and
P,(V a +0)=P*(V h +0), or p,=p h (. . . (491)
Equating (488) to (490),
* = '> or
;and
p _P b V h +P h O
Ve +
Equating (489) to (491)
P t V*+PeO T>(V+0\ PV4pnp(v
(y^oT = A, or P t V t +P e OP h (V h
and
Therefore D _
JLe
Ve+0
P t ,V b O+P h 2 =P h (V h +0)(Ve+0)P t V t (Ve+0);
V h Vc+V e O+V h O '
328
Therefore
ENGINEERING THERMODYNAMICS
n PV b O+P k V k (V c +0)^ . (fl)
Substitution will give
Pe
(V e +0)(V h +0)0 2
"^7 6 0+P t F t (F e +0)l [^0]
* T7/T7 L^^y e Q [V + 0\
(6)
(c)
_ rpy(7+Q)+p,7 t o] [
d ~ 0 2 J I
. (492)
It will be found that the use of these pressures is equivalent to the applica
tion of the equation of condition given in Eq. (487), for substitution of them
reduces to an identity, therefore the work of the two cylinders can be set down
by inspection in terms of point pressures and volumes and the above pressures
substituted. The result will be the work in terms of the pressures and cylinder
dimensions.
= P 6 7 6 (l+loge ^) PeVePeV e loge ^PeF e O ge ^
Therefore
!FjrPF(l+log, ^) 
J \ ' O,
(4: '
P.V, log,
Therefore
log, P*F t log, 
t rp & 7 ft o+p t y,(Fe+Q)
(494)
WORK OF PISTON ENGINES
329
Adding W H and W L
W =
L(^ + 0)(F + 0)
V(
(495)
F/
While this Eq. (495) for the cyclic work is in terms of initial data, it is not
of very much value by reason of its complex form. To show more clearly
i that only primary terms are included in it, the substitution of the usual symbols
will be made.
TF=144X
(bk.pr.) (X L +c L )D L log e ^^
CL
(in.pr.)c//D#  (bk.pr.) (1 X L )D L
(w^pr^ZH +CH)DH[(Z L +c L }D L +0] + (bk.pr.) (X L +c^)Dz,0
+0}[(Z L + CL )D L +0]0*
tf+ca)i>H+Ql
[(X a +C H )DH +0]
(in.pr.) (Z H +c H )D H + (bk.pr.
I
(Z L +CL)DL[(X H +CH)DH
\(Z L +c L )D L +0~]
c L D L +0
(in.pr.) (Z H +c H }D H + (bk.pr.) [(X L
(Z L +c L )D L
H +0] + (X H +c H )D H
X
X
X
. (496)
Such equations as this are almost, if not quite, useless in the solution of
problems requiring numerical answers in engine design, or in estimation of engine
performance, and this fact justifies the conclusion that in cases of finite receivers
graphic methods are to be used rather than the analytic for all design work. When
^estimates of power of a given engine are needed, this graphic work is itself seldom
Justifiable, as results of sufficient accuracy for all practical engine operation
problems can be obtained by using the formulas derived for infinite receiver when
'reasonably, large and zero receivers when small and the pistons move together.
330
ENGINEERING THERMODYNAMICS
It might also be possible to derive an expression for work with an equivalent
constantreceiver pressure, that would give the same total work and approxi
mately the same work division as for this case, but this case so seldom arises that
it is omitted here.
Inspection of the work equations makes it clear that any attempt to find
equations of condition for equal division of work for the general case must be
hopeless. It is, however, worth while to do this for one special case, that of
complete expansion and compression in both stages, yielding the diagram Fig.
98. This is of value in drawing general conclusions on the influence of receiver
size by comparing with the similar case for the infinite receiver.
By referring to Fig. 98, it will be seen by inspection that cylinder sizes,
clearances and events of the stroke must have particular relative values in order
FIG. 98. Special Case of Cycles XIII and XIV, Complete Expansion and Compression in
both Cylinders of Compound Engine with Clearance and Finite Receiver.
to give the condition assumed, i.e., complete expansion and compression. It!
is, therefore, desirable to state the expressions for work in terms which may
be regarded as fundamental. For this purpose are chosen, initial pressure
(in.pr.), back pressure (bk.pr.); highpressure displacement, D H ] cylinder
ratio, RC] highpressure clearance, C H ; and ratio of receiver volume to high
pressure displacement, y. Call
bk.pr.
= n, P .
It will be convenient first to find values of maximum receiver pressure
(rec.pr.)i, and minimum (rec.pr.) 2 ; highpressure cutoff Z H , and compressioi
X H ; lowpressure clearance CL, cutoff Z L , and compression, XL, in terms o
these quantities. Nearly all of these are dependent upon the value of CL am
it will, therefore, be evaluated first.
WORK OF PISTON ENGINES 331
From the points C and /, Fig. 98,
From A and E,
(rec.pr.)i = (in.pr.)^ , (498)
RcCi!
and from E and C,
(rec.pr.)
(rec.pr.) 2 R c c L +y '
Dividing Eq. (498) by Eq. (497) and equating to Eq. (499),
C H ) = 1+Cn+y
Multiplying out and arranging with respect to C L , the relation to be
fulfilled in order that complete expansion and compression may be possible is,
= Q. (500)
This is equivalent to
n = 0, ....... (501)
and the value of CL is
(502)
It is much simpler in numerical calculation to evaluate I, m, and n and
insert their values in Eq. (502) than to make substitutions in Eq. (500), which
would make a very cumbersome formula. (rec.pr.)2 and (rec.pr.)i may
now be evaluated from Eqs. (497) and (498) by use of the now known value of C L .
Highpressure cutoff ZH may be found from the relation of points B and
J, Fig. 98,
tip
Lowpressure cutoff, Zi L from,
or
ZH= ^(I+ CL ) CH (503)
or
Z L J^L CL (504)
tic
Highpressure compression, X H ,
C H (505)
332 ENGINEERING THERMODYNAMICS
Lowpressure compression, X L , by the use of points A and K t
Rc(X L +c L )=R P c H ,
or
X L = ^C H C L ......... (506)
He
If C L is regarded as being part of the original data, though it is related to
R c , R P , C H and y as indicated in Eq. (488), the expressions for high and lowpres
sure work and may be stated as follows:
= 144(in.p r O
. (508)
Adding these two equations gives the total work of the cycle as follows:
W= 144(in.pr.)Z>*ftc(l+c t ) f i [l+log, f? ^1 +~ log,
(ttp\_ n:c(l+C/JJ /tp
1+cu+y } g . H }
" ' 10ge
log*
This, however, may be greatly simplified,
logc
and
10gc C H ' "* e R c c L
Hence
R P .
(599)
WORK OF PISTON ENGINES
333
From this may be obtained mean effective pressure referred to the low
pressure cylinder, work per cubic foot supplied, and consumption per hour
per indicated horsepower, all leading to the same results as were found for the
case of complete expansion and compression with infinite receiver (Section 15,)
and will not be repeated here.
To find the conditions of equal division of work between cylinders, equate
Eqs. (507) and (508).
lich may be simplified to the form,
l+cn+y , , fi*p(l+c g ). RPCH \ } #W J _
T&I^ R^'
(510)
_.iis equation reduces to Eq. (376) of Section 11, when CH and CL are put
equal to zero. In its present form, however, Eq. (510) it is not capable of
solution, and it again becomes apparent that for such cases the graphical solution
of the problem is most satisfactory.
Example 1. Method of calculating Diagram, Fig. 97.
Assumed data:
P a =P b = 120 Ibs. per square inch abs.
p m =p h = 30 Ibs. per square inch abs.
PJ= 10 Ibs. per square inch abs.
To locate point C:
To locate point M :
P h V h 120 X. 4
30
Vj=V t = 2 cu.ft.
V d = Vt= 8 cu.ft.
V g = Vi= .24 cu.ft.
y e = .2 cu.ft.
V a = Vf= .12 CU.ft.
y =i.2 cu.ft.
F 6 = .4 cu.ft,
60 Ibs. per sq.in.
= 1.6 cu.ft.
To locate point D:
Pa(V d +0)=P m (V m +0}, or
= 42 Ibs. per sq.in.
334 ENGINEERING THERMODYNAMICS
To locate point E:
2 8
P e = (V e +0)=P m (V m +0), or Pe =^^3X30 =60 Ibs. per sq.in.
To locate point F:
To locate point L:
To locate point N:
p =
K/
since P n =Pe
To locate pcint G:
P a (V +0) =Pn(V n +0] or, P^O^;;; =55.5 Ibs. per sq.in.
To locate point H:
or
V, . .24X55.5+55.5X1.230X1.2 _ ^ ^ ft
To locate point /:
Example 2. Find the horsepower of a 12 and 18x24in. engine, running at 125
R.P.M., with a receiver volume twice as large as the lowpressure cylinder, 6 per cent
clearance in the highpressure cylinder, 4 per cent in the low, when the initial pressure
is 150 Ibs. per square inch absolute, back pressure 10 Ibs. per square inch absolute, high
pressure cutoff \ t lowpressure f , highpressure compression 10 per cent, low 3 per cent.
From Eq. (496) divided by 144Z>z,, and with the values as given above, the (m.e.p.)
is equal to following expression:
150 X .56[.79 X2.25 +4.5] +10(.34)2.25 X4.5 1 .16 1.06X2.25+4J
[.16+4.5][.79x2.25+4.5](4.5) 2 2.25 ge .06 X .16 X
150 X .56 X4.5 + 10 X .34 X2.25[.16 +2.25] ".79x2.25+4.5 .79X2.25+4J
]
J''
.79 X2.25[.16 +2.25] + (.16 X2.25) [04 X2.25 +4.5J ' 4 X2 ' 25 loge .04 X2.25 +4.;
150X.56X4.5+10(.34X2.25)[.16+2.25][V 1.041
f .79x2.25[.16+4.5] + (.16x2.25) [ ^ ^~J =4 ' 2 lbS " ^ ^^
hence the horsepower will be 191.
WORK OF PISTON ENGINES 335
Prob. 1. Find the work done in the highpressure cylinder and in the lowpressure
cylinder of the following engine under the conditions given.
Engine 14 and 30x28 ins., 100 R.P.M., 5 per cent clearance in each cylinder,
highpressure cutoff i 3 o, lowpressure cutoff T 4 o, highpressure compression TO, low
pressure compression TO, initial pressure 100 Ibs. per square inch gage, back pressure
5 Ibs. per square inch absolute, and receiver volume 3 times the highpressure
displacement . Logarithmic expansion.
Prob. 2. The following data are available: initial pressure 200 Ibs. per square inch
absolute, back pressure 10 Ibs. per square inch absolute, engine 10x15x22 ins., with
5 per cent clearance in the high and lowpressure cylinders, speed 100 R.P.M. What will
3e the cutoffs, and compression percentages to give complete expansion and compression.
Logarithmic expansion?
Prob. 3. What will be the work done by the above engine working under these
conditions?
Prob. 4. What must be the lowpressure clearance, cutoffs, and compression
[percentages, to give complete expansion and compression for a similar engine work
ing under the same conditions as those of Prob. 2, but equipped with a receiver twice
!is large as the highpressure cylinder?
18. Compound Engine with Finite Receiver, Exponential Law, with
Clearance and Compression, Cycle XIV. General Relations between
[Pressures, Dimensions, and Work when H.P. Exhaust and L.P. Admission
lire Independent. It cannot be expected that the treatment of this cycle by
formulas will give satisfactory results, since even with the logarithmic expansion
w, Cycle XIII gave formulas of unmanageable form. For the computation of
work done during the cycle, however, and for the purpose of checking pressures and
work determined by graphical means, it is desirable to have set down the relations
dimensional proportions, initial and final pressures, and valve adjustments, to
receiver pressures, release pressures and work of the individual cylinders.
The conditions of a steady state, explained previously, require that (Fig. 97)
firhich is the same as to say, that the quantity of fluid passing per cycle in the
lighpressure cylinder must equal that passing in the low. Expressed in terms
if dimensions,
i
 n 7? ( . 7 }\ ( cu ^"ff prQ^I _
( in.pr.)
>r, rearranging, and using R P = rrr^ N
(DK.pr.;
tm^r^H ~rnc \i<Li~r.t* LI "77 i \^n i ^*HJ \ /
R P  ,_ (m.pr
j.
(in.pr.) J i
4_7? t r 7 x[" (cutoff pr.)z,] x. 12 v
+#cfe+^ ( . npr) J , .(512)
336 ENGINEERING THERMODYNAMICS
an equation which contains two unknown pressures (rec.pr.)i and (cut off pr.) L
To evaluate either, another equation must be found:
i
/p k \
' n = Ptl T
where P n = P c , so that
Hence
v +
or
 (, , Y J(bk.pr.) I.
fo+^
(cutoff pr.) t = (recpr.)!
_ _ a
[ y (rec.pr.) * i + R c (c L +X L ) (bk.pr.) 1 '
L y+flcfo+Zi) J, V '
which constitutes a second equation between (cutoff pr.)z, and (rec.pr.)i, which
used with Eq. (512) makes it possible to solve for the unknown. By substitu
tion in Eq. (512) and rearranging,
ir
(rec.pr.)i bk.pr.) rcLciL
This expression is of great assistance even in the graphical constructio
of the diagram, as otherwise, with all events known a long process of trial an
error must be gone through with. It should also be noted that when s = l th
expression does not become indeterminate and can, therefore, be used to solv
for maximum receiver pressure for Cycle XIII, as well as Cycle XIV.
Cutoff pressure of the lowpressure cylinder, which is same as the pressure L
H or at M, Fig. 97, is now found most easily by inserting the value found b
Eq. (515) for (rec.pr.)i in Eq. (514).
Enough information has been gathered now to set down the expressions fc
work.
WORK OF PISTON ENGINES 337
. . (517)
Addition of these two Eqs. (516) and (517) gives an expression for the total
work TF, and equating them gives conditions which must be fulfilled to give
equality of work in the high and lowpressure cylinders. Since these equations
I so obtained cannot be simplified or put into more useful form, there is no object in
inserting them here, but if needed for any purpose they may be easily written.
In finding the conditions of equal work, the volumes of (rec.pr.)i and (cutoff pr.)^
must be inserted from Eqs. (514) and (515) in (516) and (517), in order to have
terms in the two equations consist of fundamental data. This, however, increases
greatly the complication of the formula.
After finding the total work of the cycle, the mean effective pressure referred
to the low pressure is obtained by dividing by 144XD.L.
To assist in finding the work per cubic foot supplied and consumption,
and the cubic feet or pounds per hour per I.H.P. it is important to know the
volume of fluid supplied per cycle,
i
(rec
ec.pr.)A "]
E^F.)/ J
Example. Find the horsepower of and compressed air steam used by a 12 and
18x24in. engine running at 125 R.P.M., with a receiver volume twice as large as the
lowpressure cylinder, 6 per cent clearance in the highpressure cylinder, 4 per cent in
the low, when the initial pressure is 150 Ibs. per square inch absolute, back pressure
10 Ibs. per square inch absolute, highpressure cutoff , lowpressure cutoff f,
highpressure compression 10 per cent, lowpressure compression 30 per cent, and
expansion and compression follow the law PV lA =c.
From Eq. (515) (rec.pr.)i is found to be as follows when values for this problem
are substituted:
^ r)l , lo
16 [4.5 +2.25 X .79] +4.5 X2.25 X .79
and by using this in Eq. (514)
.71 .71
.
/4.5X81.7 +2.25X.34X10 \ __ ., . , .,
(cutoff pr.)z, = ( 4. 5+2 .25^79" ~ ) =53 lbs * 8q>m  absolute '
338 ENGINEERING THERMODYNAMICS
It is now possible by use of Eqs. (516) and (517) by addition and division by
144D* to obtain (m.e.p.). Substituting the values found above and carrymg out
the process just mentioned.
10x2.25X.7 =51.5 Ibs. sq.in.
hence the horsepower will be 200.
By means of Eq. (518) the supply volume may be found. This gives upon
substituting of the proper values:
(Sup.Vol.) =L
L
13,750 Sup.Vol.
Cubic feet per hour per I.H.P. = ^ ^X ^ ,
_ 13,750 .46
= 51.5 X 2.25"
hence the total volume of air per hour will be
54.5X200 = 10900 cu.ft.
Prob. 1. What will be the receiver pressure and L.P. cutoff pressure for <
crosscompound compressed air engine with 5 per cent clearance in each cylinder, run
ning on 100 Ibs. per square inch gage initial pressure and atmospheric exhaust, whei
the highpressure cutoff is i, lowpressure f, highpressure compression 15 per cent
low 25 per cent, and s = 1.4. Receiver volume is twice the highpressure cylinde
volume.
Prob. 2. Find the superheated steam per hour necessary to supply a 14 and 21 X28
in. engine with 5 per cent clearance in each cylinder and a receiver twice 1 the siz
of the highpressure cylinder when the initial pressure is 125 Ibs. per square inch gage
back pressure 7 Ibs. per square inch absolute, speed 100 R.P.M., highpressur
cutoff , lowpressure T 4 o, highpressure compression 15 per cent, low pressure 40 pe
cent and s = 1.3.
NOTE: 8 = .3.
Prob. 3. If the highpressure cutoff is changed to I without change of any othe
factor in the engine of Prob. 2, how will the horsepower, total steam per houi
and steam per horsepower per hour be affected? If it is changed to f ?
Prob. 4. A boiler capable of supplying 5000 Ibs, of steam per hour at rated load fur
nishes steam for a 12 and 18 x24in. engine with 5 per cent clearance in each cylinder an*
running at 125 R.P.M. The receiver is three times as large as the highpressure cylindei
WORK OF PISTON ENGINES
339
the initial pressure 150 Ibs. per square inch gage, back pressure 5 Ibs per square inch
absolute, the lowpressure cutoff fixed at \ and lowpressure compression fixed at
30 per cent. At what per cent of its capacity will boiler be working for these follow
ing cases, when =1.2 for all and 20% of the steam condenses during admission ?
(a) highpressure cutoff I, highpressure compression 80 ] er cent,
(6) highpressure cutoff \, highpressure compression 20 per cent,
(c) highpressure cutoff , highpressure compression 10 per cent.
NOTE: = .33.
INDICATOR CARDS OF EQUAL BASE
AND HEIGHT
P
A' A
\
(in.pr.)
(rel.pr.) H
(in. pr.) L
QH
Pr.)i
X L D
alz
E"
K
I
l+CH+KcC.1
Rc1
(bk.pr.)
IM
r, ri+CH+RcCul
" DL L Rci J
FIG. 99. Work of Expansion in Compound Engine without Receiver and with Clearance.
Cycle XV, Logarithmic Expansion; Cycle XVI, Exponential, Highpressure Exhaust
and Lowpressure Admission Coincident.
19. Compound Engine without Receiver. Logarithmic Law, with Clear
iance and Compression, Cycle XV. General Relations between Pressures,
'Dimensions, and Work when H.P. Exhaust and L.P. Admission are Coinci
; jdent. The graphical construction for this cycle has been described to some
extent in connection with the first description of the cycle, given in Section 8, of
.this chapter, and is represented here by Fig. 99 in more detail.
To show that the expansion from D to E is the same as if volumes were
bieasured from the axis ML, consider a point Y on DE. If the hypothesis
is correct
(519)
340 ENGINEERING THERMODYNAMICS
The true volume when the piston is at the end D of the stroke, i:
D H (l+c a +RcCL), and at Y, the true volume is
D a (l+c a +Rcc L )D H y+D L y,
where y is the fraction of the return stroke that has been completed in bo
cylinders when the point Y has been reached. Then
P d Dn(l + CH +RCCL) = P V D H (\ +c H +R c c L ) +P v D H (Rc+l}y.
Dividing though by (R c 1),
 1 . (520)
J
__This equation may be observed to be similar in form to Eq. (519). More
over, the last term within the bracket, D H y, is equal to the corresponding tern
Y'K, in Eq. (519), hence,
, 01
' ' ' ' (521
Similarly, the distance QM, or equivalent volume at D' is
. . ..." (522
The following quantities will be evaluated preparatory to writing the expra
sions for work:
v
...... (522
, (bk.pr.)
SC
(in.pr.)i=P' = (rel.pr.)ff
 . 1
1/7 . T Hctip  T ^  ; 
Z "+ C ^ 1 y (m.pr.) Z tf +c H J
l
1fc// _ __ . .. . 4
 N (bk.pr.)
\K c (&LrCL)('
(52.
(cutoff p,),= (in. P ,), ^
WORK OF PISTON ENGINES
The ratio of expansion from E' to G is equal to
(cutoff pr.)z, I+CL
341
(rel.pr.)z,
(527)
Hence
V H = 144D,,(m.pr.) { (Z H +c H ) \l +log e

ZB+CH\
Z H +c H +Rc(X L +cz,)
(m.pr. )
Rcl
(bk.pr.)"
(CH+XH) loge CH
CH
(528)
\\z
Jn.pr.) _^
x
1+C.
CL )
. . . (529)
The total work found by adding W H and W L as given above, leads to the
following :
^(in.pr.) j (ZH+CH) Iog 6 ( 
m.pr.
144Dx,(bk.pr.) j 1 
loge
. . . (530)
342 ENGINEERING THERMODYNAMICS
This is the general expression for the work of the ^compound engine without
receiver, with clearance and compression, when highpressure exhaust and low
pressure admission are simultaneous and expansion and compression logarithmic,
in terms of fundamental data regarding dimensions and valve periods.
From this the usual expressions for mean effective pressure, work per
cubic foot supplied, and consumption per hour per I.H.P., may be easily
written, provided the supply volume is known. This is given by
(Sup. Vol.) =A?B = D H [(Z a +c a ) 
. . (531)
To find the conditions which must be fulfilled to give equal work in the two
cylinders, equate Eqs. (528) and (529).
0. (532
These expressions are perfectly general for this cycle, and expressed ii
terms of fundamental data, but are so complicated that their use is ver
limited, as in the case of some of the general expressions previously derivec
for other cycles.
As in other cycles, it is desirable to investigate a special case, that o
complete expansion and compression in both cylinders, Fig. 100. First it i
necessary to determine what are regarded as fundamental data in this case, an
then to evaluate secondary quantities in terms of these quantities. The followin
items are assumed to be known: (in.pr.), (bk.pr.), which is equal to (rel.pr.)j
R c , C H , and C L , and D H , which are dimensions, and it is known that tb
pressure at the end of compression in L.P. is equal to (rel.pr.)^.
Referring to the diagram, displacements, clearances, and the axis for
common expansion, ML, can all be laid out, and the location of the poinl
A and G determined.
The points E and E' are at the end of the common expansion within tl
two cylinders, and at beginning of highpressure compression and of separal
lowpressure expansion, hence p e = pe.
WOEK OF PISTON ENGINES
From the points A and E:
frfrHOzfe
From the points G and E' ',
/V=pe = (bk.pr.)
343
M V
IG. 100. Special Case of Cycles XV and XVI. Complete Expansion and Compression in
both Cylinders of Compound Engine without Receiver and with Clearance High
Pressure Exhaust and LowPressure Admission Coincident
and equating,
Whence
[where
(533)
Substituting the value of X H in either of the expressions for p e , which is
he lowpressure cutoff pressure,
(cutoff
(534)
344 ENGINEERING THERMODYNAMICS
It may be noted here that the cylinder ratio does not enter into this, but
only clearances and pressures. In the noclearance case, it may be remembered
that the point E or E' was not present, as it coincided with G.
Next, to find the highpressure release pressure, p dj by means of points E
and Z), and their relation to the axis ML, Fig. 100.
.RC(I~^CL) \~RpCn /COK\
(bk.pr.)  ,>  (535)
Knowing the release pressure of the highpressure cylinder, it is possible tc
find the highpressure cutoff and compression necessary to give the required
performance.
,(rel.pr.) g
p I_L^ LP n
(in.pr.) HP L i+CH\ticCL J
(rel.pr.) g \
/i i ^  CL = CL\
(bk.pr.) L
(rel.pr.) g \Rc(l+c L )+RpCH J
=  = ~  1
J
The work of the two cylinders is as follows:
w
^cci,] \Rc(i+c L )+Rpc H ~\ j_ ,
l JL 1+CH+RcCL jRp ^
Re
' )
R P (R
+fipcir], rg c (i+te)
l) J * L l+c^+
I+C L + CH
WORK OF PISTON ENGINES 345
These expressions, when added and simplified, give the following for total
work per cycle,
( r r> /i i _ \ i r> _ ~~i 'i
. . (540)
in which of course D H R C may be used instead of D L and ^ ?P^ instead of
K P
(bk.pr.) and then
p, . ... (541)
Z H having the value of Eq. (536).
Equality of work the in high and lowpressure cylinders results, if W u
Eq. (538) equals W L , Eq. (539), or if
, or
all of which lead to equivalent expressions. Simplification of these expres
sions, however, does not lead to any direct solution, and hence the equations
will not be given here.
Example. Find (a) the horsepower and (6) steam used per hour for a 12 and 18 x24in.
tandem compound engine with no receiver, 6 per cent clearance in the highpressure
cylinder, and 4 per cent in the low, when the initial pressure is 150 Ibs. per square inch
absolute, back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., high
pressure cutoff , highpressure compression 15 per cent and lowpressure compression
is complete.
(a) Since the lowpressure compression is complete, the pressure at end of compression
must be equal to the release pressure of the high. This latter quantity may be found
*rom the relation (in.
/ r * Y
(rel.pr.)^ = 150 r= 79.3 Ibs. per sq.in. absolute.
1 .uo
Lowpressure compression may be found from the relation (rel.pr.)tf(cz,) = (bk.pr.)
(CL+XL), or X L = .28. (m.e.p.) may be found from Eq. (540) divided by 144Dz,, which
on substitution gives
.5+.06+2.25(.28 + .04)
+ . ...oo. ______ o 2.25U.04.15) log e
1 +.06+2.25 X.04 + 1.75+.85J * 1+.04.15
(.06 + .15) loge ~ 10 1 1.28 + (.28+.04) loge ^~^ \ =69.7. Ibs. sq. in.
.Ob J (
the horsepower will be 271.
346 ENGINEERING THERMODYNAMICS
(6) Since the consumption in cubic feet per hour per horsepower is equal to
'_13,750_ Sup.Vol.
(m.e.p.) DL
and supply volume is given by Eq. (531), this becomes
69.7 "2.251 \1.06 + .09 + 1.25X.85
hence the consumption per hour will be
44 X271 X.332 =4000 pounds.
Prob. 1. A Vauclain compound locomotive has cylinders 18 and 30x42 ins., with 5
per cent clearance in each and runs on a boiler pressure of 175 Ibs. per square inch gage
and atmospheric exhaust. The steam pressure may be varied as may also the cutoff to a
limited degree. For a speed of 200 R.P.M., a cutoff f and 10 per cent compression
in each cylinder, find how the horsepower will vary with the initial pressures of 175,
150, 125, and 100 Ibs. gage.
Prob. 2. When the cutoff is reduced to in the above engine compression
increases in the highpressure cylinder to 20 per cent For the case of 175 Ibs. gage
initial pressure find the change in horsepower.
Prob. 3. Find the steam used by the engine per hour for the first case oi
Prob. 1 and for Prob. 2.
Prob. 4. It is desired to run a 12 and 18x24in. noreceiver engine with 5 per cenl
clearance in each cylinder, under the best possible hypothetical economy conditions for ar
initial pressure of 200 Ibs. per square inch absolute and atmospheric exhaust. To
give what cutoff and compression must the valves be set and what horsepowe]
will result for 100 R.P.M.?
20. Compound Engine Without Receiver. Exponential Law, with Clear
ance and Compression, Cycle XVI. General Relations between Pressures
Dimensions, and Work, when H.P. Exhaust and L.P. Admission are Coin
cident. Again referring to Fig. 99, it may be observed that reasoning
similar to that in Section 19 but using the exponential law, would show thai
the same formulas and graphical constructions will serve to locate th(
axes of the diagram, hence, as before,
, (542
and
= DS ' rr^T 7 ^. . (543
Kc~~ 1
Release pressure in the highpressure cylinder is
)' (544
WORK OF PISTON ENGINES
347
Immediately after release the pressure is equalized in the highpressure
cylinder and the lowpressure clearance. The pressure after equalization,
termed (in.pr.) L , is found by the relation of the volume at S and that at D',
Fig. 99, measured from the axis KT.
which, by means of Eq. (544) becomes
The expansion of the fluid goes on as it passes from the highpressure
cylinder to the greater volume in the lowpressure, as indicated by D'E'
>and DE, and when the communicating valve closes, the pressure has become
(cutoff pr.) L = (in.pr.
which, by means of Eq. (545) reduces to
i
After cutoff in the low pressure, expansion goes on in that cylinder alone
to the end of the stroke, when release occurs at a pressure
(rel.pr.)z, = (cutoff pr
r by substitution from equation (546),
\1* (547)
)\ '
nr ),  ft>k r.r )
k  pr0 L l+c H +RcC L +(lX H )(R c l)
In terms of these quantities the work of the high and lowpressure
cylinders can be written out as follows:
frir = 144D*
s1
(bk.pr.)
'r / I+CH+RCCL V" 1
I \l+c H +Rcc L +(lX H )(R c l) /
+^y 1 ill /ox
7~y
348 ENGINEERING THERMODYNAMICS
and
(bk.pr.)
RCI r
x\i
\l+c H +R c c L +(lX)(Rc
s1
I
. (bk.pr.) , J R*P(CH+ZH)+RC(CL+X L } ] s ~ /l + c L X H \'^
L  .1 J^ O ^^ // ) I ~" \ ~"\" I I J. """" I I
:.pr.) . (549)
These are general expressions for work of high and lowpressure cylinders
for this cycle, and from them may be obtained the total work of the cycle,
mean effective pressure referred to the lowpressure cylinder, and by equating
them may be obtained the relation which must exist between dimensions,
events, and pressures to give equal division of work. It would, however,
be of no advantage to state these in full here, as they can be obtained from
the above when needed.
The supply volume, cubic feet per cycle, is represented by A 'B, Fig. 99,
and its value is found by referring to points B and E as follows:
(Sup.Vol.) =D a \(c a +Z a )  (c a +X a ) ((cutoff P r.)A! ,
\ (m.pr.) / J
n L 4.7 (CH+X H ) ( R> P(CH+Z H )+RC(CL+X L ) \1 . .
~T ( l+ c a +Rcc L +(lX H )(R c T)) J ' (55
Work per cubic foot supplied is found from Eqs. (548), (549), and (550).
Work per cu.ft. supplied = .. . . . . . >. . . (551)
Consumption, cubic feet per hour per I.H.P., is found from mean effective
pressure referred to L.P. cyl. and supply volume as follows:
Consumption, cu.ft. per hr. per I.H.P.
13,750 (Sup.VoL) I
(m.e.p. ref. to L.P.) D L
This will give pounds consumption by introducing the factor of density. 
Further than this, it will be found more practicable to use graphica
methods instead of computations with this cycle.
WORK OF PISTON ENGINES 349
Example. Find (a) the horsepower and (6) consumption of a 12 and 18x24in.
noreceiver engine having 6 per cent clearance in the high pressure cylinder and 4 per
cent in the low when the initial pressure is 150 Ibs. per square inch absolute, back
pressure 10 Ibs. per square inch ab.iolute, speed 125 R.P.M., highpressure cutoff ,
highpressure compression 15 per cent, and lowpressure compression is complete.
(a) The per cent of lowpressure compression may be found as in the Example of
Section 19, using the value of s in this case of 1.4. Then
or
(rel.pr.)#=61.5 Ibs. sq. inch absolute,
and
or
!.
From the sum of Eqs. (548) and (549) divided by 144Z>z, and with proper
alues substituted, (m.e.p.) =48.5 Ibs.; hence the horsepower is 189.
13 750
(b) From Eqs. (550) the value for (Sup. Vol.), which when multiplied by  ,
m.e.p.
and divided by D L , gives cubic feet air per hour per I.H.P.
!LZ?y JF
48.5 2.25[
15 7 (.56)+2.25(.15) _] =. 63 cu . ft . per
(15)' 7 1 +.06+2.25 X.04 + (l .15)(2.251) hourper I.H.P.
Prob. 1. If the locomotive of Prob. 1, Section 19, should be equipped with superheater
o that the steam expanded in such a way that s = 1.3, what would be the effect upon
tie horsupower for conditions of that problem and on the cylinder event pressures?
Prob. 2. A 30 and 42x54in.no receiver steam pumping engine runs at 30 R.P.M.
nd has 3 per cent clearance in the highpressure cylinder and 2 per cent in low. There
is no compression in either cylinder. Initial pres ure is 120 Ibs. per square inch gage,
md back pressure 28 ins. of mercury (barometer reading 30 ins.). The steam is such
bat the expansion exponent is 1.25. What will be the horsepower of, and the steam
ised by the engine when the cutoff in the high is ^?
Prob. 3. By how much would the power change if the cutoff were shortened to f
and then to i, and what would be the effect of these changes on the economy?
21. TripleExpansion Engine with Infinite Receiver. Logarithmic Law.
So Clearance, Cycle XVII. General Relations between Pressures, Dimen
sions and Work. Fig. 101 represents the cycle of the tripleexpansion engine
with infinite receiver, no clearance, showing one case of incomplete expansion
p. all cylinders, and another where overexpansion takes place in all cylinders.
350
ENGINEERING THERMODYNAMICS
The reasoning which follows applies equally well to either case, and to an;
combination of under or overexpansion in the respective cylinders.
It is desired to express the work of the respective cylinders and th
total work in terms of dimensions, initial and back pressures, and the cutoff
of the respective cylinders. To do this, it is convenient first to express th
B E F
H.P.V I. PA
\ D \
(1st rec.pr) INDICATOR CARDS OF EQUAL
INCOMPLETE EXPANSION
V
CYCLE XVII
OVER EXPANSION
FIG. 101. Work of Expansive Fluid in TripleExpansion Engine with Infinite Receiver i
Zero Clearance. Cycle XVII, Logarithmic Expansion.
first receiver pressure (1st rec.pr.) and second receiver pressure (2d rec.pr.)
terms of these quantities. The subscript / refers to the intermediate cylind
or
and
/i j_ \ / \ ^HL) ff
(1st rec.pr.) = (in.pr.) " ,
rFi
(55
or
(2d rec.pr.) = (in.pr.)^^.
(55
WOEK OF PISTON ENGINES
Work of highpressure cylinder is
351
144(in.pr.)A,
Work of intermediate cylinder is
(555)
^jf~4 (556)
Work of lowpressure cylinder is
. . . (557)
The total work by addition is
D
*&\uw****
, . (558)
Mean effective pressure referred to the lowpressure cylinder is found by
[dividing W by 144D/,, and is therefore
i.e.p. ref. to L.P.)
352 ENGINEERING THERMODYNAMICS
Work done per cubic foot supplied is equal to W divided by the supp
volume Z# or Z H D H ,
Work per cu.ft. supplied
= 144(in.pr.) ( 3 +log c ~jr % ~  144(bk.pr.)TF . (5ft
LH^I^L AiDi ALL)L} &HL)H
The volume of fluid supplied per hour per indicated horse power is
Consumption, cu.ft. per hr. per I.H.P.
13,750 Z H D
(m.e.p. ref. to L.P.) D L
(56
The weight of fluid used per hour hour per indicated horsepower is
course found by multiplying this volume Eq. (561) by the density of the flu
used.
The conditions which will provide for equal division of work between tl
three cylinders may be expressed in the following ways:
which is equivalent to, first:
or
'Oge  1/A
hence
Similarly from
These two equations, (562) and (563), show the necessary relations betf 
7 7 7 ( Dl \ ( DL \ /(in.pr.)\
Ztf, Z 7 , Zz,, (7^), {FT , and 7^^41,
\D*I WJT/ \(bk.pr.)/'
in order that work shall be equally divided. Since there are six independ
quantities entering (as above) and only two equations, there must be }
of these quantities fixed by conditions of the problem, in order that the ot
WORK OF PISTON ENGINES
353
may be found. For instance, if the cylinder ratios, the pressure ratio,
and one cutoff are known, the other two cutoffs may be found, though the
solution is difficult.
Again, if cutoffs are equal, and the ratio of initial to back pressure is known,
it is possible to find the cylinder ratios. This forms a special case which is of
sufficient importance to require investigation.
If Zn = Zi = Z L , Eq. (562) becomes
D
and Eq. (563) reduces to
but from Eq. (564),
and therefore
and
D I D L (m.pr.)
(bk.pr.)'
D 2 DH DH \D H '
D
D
which, along with the condition assumed that
Z H = Zj =
(564)
(565)
(566)
(567)
lonstitute one set of conditions that will make work equal in the three cylinders
frhis is not an uncommon method of design, since by merely maintaining
equal cutoffs, the work division may be kept equal.
The work done in any one cylinder under these conditions Eqs. (566)
ind (567) is then
nd the total work
 (568)
which Z represents the cutoff in each cylinder, all being equal.
354
ENGINEERING THERMODYNAMICS
A special case of the tripleexpansion engine with infinite receiver and n<
clearance which demands attention is that of complete expansion in al
cylinders, represented by Fig. 102. Here
D t _ (bk.pr.)
_> /y=: (2d rec.pr.)
' Z), (1st rec.pr.)'
FIG. 102. Special Case of Cycle XVIII Complete Expansion in Tripleexpansion Engii
with Infinite Receiver, Zero Clearance, Logarithmic Expansion.
and
7 _ (1st rec.pr.) _ /bk.pr A D L
(in.pr.) UprjA/
hence the receiver pressures are as follows:
(1st rec.pr.) = (bk.pr.;
(57:
WORK OF PISTON ENGINES 355
(2drec.pr.)=(bk.pr.)^ ........ (574)
The work of the respective cylinders, expressed in terms of initial and
)ack pressures and displacements is then,
,,,
Similarly,
e \J)ff/ } w'w
and
Total work, by addition, is
(57g)
If for this special case of complete expansion equality of work is to be
jfctained, then from Eqs. (575), (576), and (577),
(in.pr.) D H = D I= D L
(bk.pr.) D L D H DS
Bich is readily seen to jbe the same result as was obtained when all cutoffs
lere equalized, Eqs. (564) and (565). This case of complete expansion and
ual work in all cylinders is a special case of that previously discussed where
fctoffs are made equal. Hence for this case cutoffs are equal,
777 _K_I_L.PT.
^
Example. A tripleexpansion engine 12 and 18 and 27x24ins., with infinite
ceiver and no clearance, runs at 125 R. P.M. on an initial pressure of 150 Ibs.
r square inch absolute, and a back pressure 10 Ibs. per square inch absolute.
356 ENGINEERING THERMODYNAMICS
If the cutoffs in the different cylinders, beginning with the high, are J, , and I,
what will be (a) the horsepower, (6) steam consumed per hour, (c) release and
receiver pressures?
(a) From Eq. (559)
(bk.pr.),
hence
'39X2X573X250
33,000
(6) From Eq. (561).
13,750 Z H D H ,
Cubic feet per horsepower per hour = f N
(m.e.p.) D L
hence total pounds per hour will be,
34.9 X.338X.332 =3920.
(c) From Eq. (553)
1st (rec.pr.) = (in.pr.)^^,
= 150 X ' =89 Ibs. per sq.in absolute.
From Eq. (554)
2d (rec.pr.) =(in.pr.
= 150 X ' =3.75. per sq.in absolute.
.0/0 Xo.uo
Highpressure release pressure may be found from relation (in.pr.) ZnDu
(rel.pr.)//D# .*. (rel.pr.)#=75 Ibs. Similarly 1st (rec.pr.)Z/D/ = (rel.pr.)/D/, 01
(rel.pr.)/=33.4. Similarly 2d (rec.pr.)Z L D L = (rel.pr.)iZ>i,, or (rel.pr.)i = 14.8.
Prob. 1. What would be the horsepower and steam used per hour by a 10 am
16 and 25x20in. infinite receiver, noclearance engine, running at 185 R.P.M
on an initial pressure of 180 Ibs. per square inch gage and atmospheric exhaust
Cutoffs .4, .35, and .3.
Prob. 2. The following data are reported for a test of a triple engine:
Size 20x33x52x42 ins., speed 93 R.P.M., initial pessure 200 Ibs. per square incl
gage, back pressure one atmosphere, H.P. cutoff .5, horsepower 1600, steam per hsore
WORK OF PISTON ENGINES 357
ower per hour 17 Ibs. Check these results, using cutoffs in other cylinders to give
pproximately even work distribution.
Prob. 3. What change in cylinder sizes would have to be made in the above engine
3 have equal work .with a cutoff of in each cylinder, keeping the high pressure the
ime size as before?
Prob. 4. What would be the horsepower of a tripleexpansion engine whose low
ressure cylinder was 36x3 ins., when running on 150 Ibs. per square inch absolute
litial pressure and 10 Ibs. per square inch absolute back pressure, with a cutoff in
ach cylinder of .4 and equal work distribution? Make necessary assumptions.
Prob. 5. A triple engine 18x24x36x30 ins., running at 100 R. P.M. on an initial
ressure of 200 Ibs. per square inch absolute and back pressure of 20 Ibs. per
quare inch absolute, is to be run at such cutoffs as will give complete expansion in
11 cylinders. What will these be, what receiver pressures will result, what horse
;ower can be produced under these conditions, and how much steam will be needed
'er hour?
NOTE: 8 for 200 Ibs. = .437.
22. MultipleExpansion Engine. General Case. Any Relation between
pylinder and Receiver. Determination of Pressure VolumeDiagram and
iVork, by Graphic Methods. It is possible to arrange multipleexpansion
jngines in an almost infinite variety of ways with respect to the pressurevolume
Shanges of the fluid that take place in their cylinders and receivers. There
pay be two or three cylinder compounds of equal or unequal strokes, pistons
loving together by connection to one piston rod, or separate piston rods
dth a common crosshead or even with completely independent main parts
nd cranks at 0, 180, displaced with either one leading, or the pistons may
ot move together, being connected to separate cranks at any angle apart,
ad any order of lead. Moreover, there may be receivers of large or small
ie, and there may be as a consequence almost any relation between H.P.
.scharge to receiver and lowpressure receipt from it, any amount of fluid
assing to correspond to engine load demands and consequently any relation
cutoffs, compressions, and receiverpressure fluctuations. Triple and
uadruple engines offer even greater varieties of combination of related factors,
i that problems of practical value cannot be solved by analytical methods
anything like the same ease as is possible by graphic means, and in
ime cases not at all.
These problems that demand solution are of two classes.
1. To find the work distribution and total work for cylinders of given
imensions, clearances, receiver volumes and mechanical connection or move
Lent relation, with given initial and back pressures, and given valve gear at
setting of that valve gear or at a variety of settings.
2. To find the cylinder relations to give any proportion of the total work
any cylinder at any given valve setting or any fraction of initial pressure
any value of release pressure or total number of expansions.
The essential differences between these two classes of problems is that in the
t the cylinder dimensions are given, while in the second they are to be found.
358 ENGINEERING THERMODYNAMICS
In general, however, the same methods will do for both with merely
change in the order, and in what follows the dimensions of cylinder
valve periods, receiver volume, initial and back pressures will be assume
and the diagrams found. By working to scale these diagrams will give tt
work by evaluation of their area, by means either of crosssection paper directb
on which strips can be measured and added, or by the planimeter. Thi
will high and lowpressure work be evaluated through the footpound equivi
lent per square inch of diagram, and the total work or the equivalent mea
effective pressure found by the methods of mean ordinates referred to th
pressure scale of ordinates.
In the finding of the pressurevolume diagram point by point there is but on
common principle to be applied, and that is that for a given mass the produc
of pressure and volume is to be taken as constant (for nearly all steam prot
lems, which is the almost sole application of this work) and when two masst
come together at originally different pressures and mix, the product of the resul
ing pressure and the new volume, is equal to the sum of the PV products of th
two parts before mixture. At the beginning of operations in the highpressur
cylinder, a known volume of steam is admitted at a given pressure and i1
pressure and volume are easily traced up to the time, when it communicate
with the receiver in which the pressure is unknown, and there difficulty
encountered, but this can be met by working from the other end of tr.
series of processes. The lowpressure cylinder, having a known compressio
volume at the back pressure, there will be in it at the time of opening
receiver a known volume, its clearance, at a known compression pressur
The resulting receiver pressure will then be that for , the mixture. These
receiver pressures are not equal ordinarily, but are related by various compre
sions and expansions, involving high and lowpressure cylinder parti
displacements, grouped with receiver volumes in various ways.
Take for an illustrative example the case of a twocylinder, singleactin
crosscompound engine with slide valves, cylinder diameters 12f and 20 in
with 24 ins. stroke for both. Highpressure clearance is 10 per cent,
pressure clearance 8 per cent. Receiver volume 4000 cu.ins. Highpressui
crank following by 90. Find the mean effective pressure for the high and k
pressure cylinders, for a cutoff of 50 per cent in the high, and 60 per ce
in the low, a compression of 10 per cent in the high and 20 per cent in the lo
initial pressure 105 Ibs. per square inch gage, back pressure 5 Lbs. p
square inch absolute, expansion according to logarithmic law.
On a horizontal line SZ, Fig. 103, lay off the distances
TU = lowpressure cylinder displacement volume in cubic inches to scale.
U 7= lowpressure cylinder clearance volume in cubic inches to scale.
VW = receiver volume in cubic inches to scale.
WX = highpressure cylinder clearance volume in cubic inches to scale.
XY = highpressure cylinder displacement volume in cubic inches to scale
WORK OF PISTON ENGINES
359
Through these points draw verticals produced above and below, T'T",
V'U", V'V", W'W", X'X", Y'Y". Then will W'W and WZ be PV
coordinates for the high pressure diagram in the quadrant W'WZ, and V'V
'IG. 103. Graphical Solution of Compound Engine with Finite Receiver and with Clearance
Illustrating General Method of Procedure for any Multiple Expansion Engine.
id FS the PV coordinates for the low pressure diagram in the reversed
ladrant F'FS.
Lay off AB to represent the highpressure admission at a height XA rep
360 ENGINEERING THERMODYNAMICS
resenting absolute initial pressure; lay off LM at a height TL representing
lowpressure exhaust at a constant absolute_back pressure to the same scale.
Locate point B at the cutoff point AB = .5QXY on the initial pressure
line, and drop a vertical BB 2 and draw similar verticals JJ 2 , GG 2 , MM 2 , at
suitable fractional displacements to represent L.P. cutoff, H.P. and L.P.
compression volumes resprectively.
This operation will fix two other points besides the points A and L, B the
H.P. cutoff at the initial pressure and M the lowpressure compression at the
back pressure. Through the former draw an expansion line BC and through
the latter a compression line MN, locating two more points, C and N, at the
end of the outstroke of the high and instroke of the low.
At point C the H.P. cylinder steam releases to the receiver of unknown
pressure, and at N, the L.P. cylinder steam is opened to both the receiver and
highpressure cylinder at unknown pressure and volume.
To properly locate these pressures and volumes from the previously known
pressures and volumes hi a simple manner, the construction below the line
SZ is used.
Lay off on W'W" the highpressure crank angles 0360, and to the right
of each lay off from the clearance line XX " the displacement of the piston at the
various crank angles for the proper rod to crank ratio, locating /the curve
A 2 B 2 C 2 E 2 F 2 G 2 H 2 . This is facilitated by Table XIII at the end of the
chapter, but may be laid out graphically by drawing the crank circle and
sweeping arcs with the connecting rod as radius.
Opposite H.P. crank angle 270 locate L.P. crank angle = 360 and
draw to left of the lowpressure clearance line UU" the crank angle dis
placement curve for that piston.
It will be noted that steam volumes are given in the lower diagram by the
distances from either of these curves toward the other as far as circum
stances call for open valves. Thus H.P. cylinder volumes are distances from
the H.P. displacement curve to WW", but when H.P. cylinder is in communi
cation with receiver, the volume of fluid is the distance from H.P. displacement
curve to VV", and when H.P. cylinder, receiver and L.P. cylinder are all
three in communication the volume is given by the distance from H.P.
displacement curve to L.P. displacement curve. This pair of displacement
curves located one with respect to the other as called for by the crank angle
relations, which may be made to correspond to any other angular relation,
by sliding the low up or down with respect to highpressure curve, serve
as an easy means of finding and indicating the volumes of fluid occupying
any of the spaces that it may fill at any point of either stroke.
On each curve locate the points corresponding to valve periods by the
intersection of the curve with verticals to the upper diagram, such as BB 2 .
These points being located, the whole operation can be easily traced.
At H.P. cutoff (B) the volume of steam is B 3 B 2 . During H.P. expansion
(B to C) steam in the high increases in volume from B 3 B 2 to C 3 C 2 .
During H.P. release (C to D) the volume of steam in the high C 3 C 2 is
WORK OF PISTON ENGINES 361
m
added to the receiver volume C 4 C 3 , making the total volume C 4 C 2 . During
H.P. exhaust (D to E) the steam volume C 4 C 2 in H.P. and rec. is com
pressed to volume I 3 E 2 .
At L.P. admission (N) in low and (E) in high, the volume I 2 ! 3 is added,
making the total volume I 2 E 2 in high, low, and receiver.
During (7 to Q) in low and (F to G) in high the volume I 2 E 2 in high, low
and receiver, changes volume until it becomes Q 2 G 2 in high, low, and receiver.
At H.P. compression, G in the high, the steam divides to Q 2 G 3 in low
and receiver, while G 3 G 2 remains in high and is compressed to H 3 H 2 , at the
beginning of admission in the high. The former volume Q 2 G 3 , in low and
receiver, expands to J 2 J 4 , at the moment cutoff occurs in the low, which
divides the volume into, J 3 J 4 in receiver, which remains at constant volume
1 till highpressure release, and the second part, J 2 J 3 in the low, which expands
[fin that cylinder to K 2 K 3 .
After lowpressure release the volume in low decreases from K 2 K 3 to
M 2 M 3 , when the exhaust valve closes and lowpressure compression begins.
During compression in low, the volume decreases from7kf 2 M 3 to I 2 ! 3 which
is the volume first spoken of above, which combines with I 3 E 2 , causing the drop
in the highpressure diagram from E to F.
The effects upon pressures, of the various mixings at constant volume
between high, low, and receiver steam and the intermediate common expan
sions and compressions may be set down as follows:
At C, steam in high, at pressure P CJ mixes with steam in receiver at
pressure Pj, resulting in high and receiver volume at pressure P d .
From D to E there is compression in high and receiver resulting in
pressure P e .
At E steam in high and receiver at pressure P e mixes with steam in low,
at P n , locating points I in low and F in high at same pressure.
From (F to G in high) and (/ to Q in low) there is a common compression
expansion in high, low, and receiver, the pressures varying inversely as the total
volume measured between the two displacement curves. At G in there begins
compression in high alone to H.
In the low and receiver from Q to / there is an expansion and consequent
fall in pressure from P a to Pj.
After lowpressure cutoff at J the expansion takes place in lowpressure
cylinder alone, to pressure P*, when release allows pressure to fall (or rise)
to exhaust pressure Pi.
When cutoff in low occurs at / the volume J 3 J 4 is separated off in the
leceiver, where it remains at constant pressure P/ until highpressure release
at point C.
tAt the point M compression in low begins, increasing the pressure in low
one from P m to P n .
There are, it appears, plenty of relations between the various inter
ediate and common points, but not enough to fix them unless one be first
established. One way of securing a starting point is to assume a compression
362 ENGINEERING THERMODYNAMICS
pressure P g for the beginning of H.P. compression and draw a compression
line HG through it, produced to some pressure line a/, cutting lowpressure
compression line at d. Then the H.P. intercept (ef) must be equal to the
lowpressure intercept (dc); this fixes (c) through which a PV = const, line
intersects the L.P. cutoff volume at /.
Now knowing by this approximation the pressure at /, the pressure may
by found at D, E, F, and at G. The pressure now found at G may differ
considerably from that assumed for the point. If so, a new assumption for
the pressure at G may be made, based upon the last figure obtained, and
working around the circuit of pressures, J, D, E, F, and back to G should
give a result fairly consistent with the assumption. If necessary, a third
approximation may be made.
It might be noted that this is much the process that goes on in the receiver
when the engine is being started, the receiver pressure rising upon each release
from the high, closer and closer to the limiting pressure that is completely reached
only after running some time.
These approximations may be avoided by the following computation,
representing point pressures by P with subscript and volumes by reference
to the lower diagram. Pj is the unknown pressure in receiver before high
pressure release and after lowpressure cutoff.
Pressure after mixing at D is then
(C 4 C 2 )
The pressure at F, after mixing is
(PE 2 )
(PE 2 }
This pressure multiplied by W gives P 8 , and this in turn multiplic
by
Writing this in full,
Solving for P h
WORK OF PISTON ENGINES 363
which is in terms of quantities all of which are measurable from the diagram.
While this formula applies to this particular case only, the manner of obtain
ing it is indicative of the process to be followed for other cases.
When there are three successive cylinders the same constructions can be
used, the intermediate diagrams taking the position of the low for the com
pound case, while the low for the triple may be placed under the high and offset
from the intermediate by the volume of the second receiver. In this case
it is well to repeat the intermediate diagram. Exactly similar constructions
apply to quadruple expansion with any crank angle relations.
Prob. 1. By means of graphical construction find the horsepower of a 12 and 18 X24
in. singleacting crosscompound engine with 6 per cent clearance in each cylinder,
if the receiver volume is 5 cu.ft., initial pressure 150 Ibs. per square inch gage, back
pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., highpressure compression
30 per cent, low pressure 20 per cent, high pressure cutoff 50 per cent, low pressure
40 per cent, highpressure crank ahead 70, logarithmic expansion, and ratio of red
to crank 4.
Prob. 2. Consider the above engine to be a tandem rather than a crosscompound
; 'and draw the new diagrams for solution.
Prob. 3. A doubleacting, 15 and 22 x24in. compound i engine has the high
pressure crank ahead by 60, and has 5 per cent clearance in the lowpressure cylinder,
10 per cent in the high, and a receiver 4 times as large as the highpressure cylinder.
What will be the horsepower when the speed is 125 R. P.M. /initial pressure 150 Ibs. per
square inch absolute, back pressure 5 Ibs. per square inch absolute, highpressure cut
off ^, lowpressure , highpressure compression 20 per cent, lowpressure 30 per cent,
and ratio of rod to crank 5. Determine graphically the horsepower in each cylinder.
Prob. 4. Consider the engine of Prob. 3 to be a tandem compound and repeat
the solution.
23. Mean Effective Pressure, Engine Power, and Work Distribution and
their Variation, with Valve Movement and Initial Pressure. Diagram Dis
tortion and Diagram Factors. Mechanical Efficiency. The indicated power
developed by a steam engine is dependent upon three principal factors piston
displacement, speed, and mean effective pressure. The first, piston displacement,
is dimensional in character, and, fixed for a given engine. Speed is limited by
steam and inertia stresses, with which the present treatment is not concerned,
or by losses due to fluid friction in steam passages, a subject that will be
further considered under steam flow. Mean effective pressure is a third factor
which is to be investigated, most conveniently by the methods laid down in
the foregoing sections.
In these formulas for mean effective pressure, it will be observed that
the terms entering are (a) initial pressure, (6) back pressure, (c) cutoff or
ratio of expansion, (d) clearance, and (e) compression, for the singlecylinder
engine. It is desirable to learn in what way the mean effective pressure
varies upon changing any one of these factors.
364 ENGINEERING THERMODYNAMICS
Referring to Section 5, Eq. (262) for logarithmic expansion
(m.e.p.) = (in. pr.) Z+ (Z+c)log c ^~ (mean forward pressure)
(bk.pr.)\l
X+(X+c)
J
(mean back pressure)
(582)
it is seen that the mean effective pressure is the difference between a mean
forward pressure and a mean back pressure. The former depends on
initial pressure, cutoff, and clearance, and the latter on back pressure,
I
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10 15 20 25 30 .1 .2 .3 A .5 .6 .7 .8 .9 1.0 .05 .10 .15 .20 .25
FIG. 104. Curve to Show Variation of Mean Forward and Mean Back Pressure for Logarithmic
Expansion and Compression in a Single Cylinder Engine with Clearance.
compression, and clearance. To study the effect of varying these terms it
is most convenient to draw curves such as are shown in Fig. 104, and examine
mean forward pressure and mean backward pressure separately.
Mean forward pressure is seen by inspection to vary in direct proportion
to initial pressure. Cutoff, when short, gives a low mean forward pressure,
but it is to be noted that zero cutoff will not give zero mean effective
pressure so long as there is clearance, due to expansion of steam in the
clearance space. Increasing the length of cutoff, or period of admission,
increases mean forward pressure, but not in direct proportion, the (m.f.p.)
approaching initial pressure as a limit as complete admission is approached.
WORK OF PISTON ENGINES
365
earance has the tendency as it increases, to increase the mean forward
pressure, though not to a great extent, as indicated by the curve Fig. 104.
Mean back pressure is usually small as compared to initial pressure, though
a great loss of power may be caused by an increase of back pressure or com
pression. Back pressure enters as a direct factor, hence the straight line through
the origin in the figure. So long as compression is zero, back pressure and
mean back pressure are equal. When compression is not zero, there must
be some clearance, and the ratio of (mean bk.pr.) to (bk.pr.) depends on both
clearance and compression, being greater for greater compressions and for
i smaller clearances,
100
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FIG. 105. Curves to Show Variation of Mean Effective Pressure for Logarithmic Expansion
and Compression in a Single Cylinder Engine with Clearance.
The mean effective pressures obtained by subtracting mean back from
mean forward pressures in Fig. 104 are shown in curve form in Fig. 105.
The multiple expansion engine can not be so simply regarded. In a general
Way each cylinder may be said to be a simple engine, and subject to variations
if mean effective pressure due to change in its own initial pressure and back
pressure, clearance, cutoff and compression, which is true. At the same time
hese factors are interrelated in a way that does not exist in the simple
gine. Consider, for instance, the highpressure cylinder of a compound
gine with infinite receiver, with clearance. An increase of highpressure
mpression tends first to raise the mean back pressure according to the
ning on simple engine, but at the same time the change has decreased
e volume of steam passing to receiver. No change having been made
366 ENGINEERING THERMODYNAMICS
in the lowpressure cylinder, the volume admitted to it will remain the same
as before, and the receiver pressure will fall, decreasing mean back pressure
by a greater amount than compression increased it, and mean forward being
the same as before, the increase of highpressure compression has increased
the mean effective pressure of the highpressure cylinder. The only effect
upon the lowpressure cylinder is that resulting from lowering its initial
pressure, i.e., the receiver pressure. This results in a decrease of lowpressure
mean effective pressure. Computation will show that the assumed increase
of highpressure compression decreases lowpressure work more than it
increases highpressure work, or in other words, decreased mean effective
pressure referred to the low.
It is impracticable to describe all results of changing each of the
variables for the multipleexpansion engine. Initial pressure and cutoff
in the respective stages have, however, a marked influence upon receiver
pressures and work distribution which should be noted. Power regulation
is nearly always accomplished by varying initial pressure, i.e., throttling,
or by changing cutoff in one or more cylinders.
The effect of decreasing initial pressure is to decrease the pressures on
the entire expansion line and 'in all no clearance cycles, to decrease absolute
receiver pressures in direct proportion with the initial pressure. Since back
pressure remains constant, the result is, for these noclearance cycles, that
the mean effective pressures of all but the lowpressure cylinder are decreased
in direct proportion to the initial pressure, while that of the lowpressure is
decreased in a greater proportion. The same is true only approximately
with cycles having clearance and compression.
The conditions giving equal work division have been treated in connection
with the individual cycles, it may here be noted in a more general way that
if highpressure cutoff is shortened, the supply capacity of that cylinder is
decreased, while that of the next cylinder remains unchanged. The result
is that the decreased supply volume of steam will be allowed to expand to
a lower pressure before it can fill the demand of the next cylinder than itl
did previously, i.e., the receiver pressure is lowered. Similarly shortening
cutoff in the second cylinder will tend to increase receiver pressure. To
maintain constant work division, there must be a certain relation between
cutoffs of the successive cylinders, which relation can only be determines
after all conditions are known, but then can be definitely computed and
plotted for reference in operation.
So far, in discussing the steam engine, cycles only have been treated
These cycles are of such a nature that they can be only approached ho
practice, but since all conclusions have been arrived at through reasoning
based on assumed laws or hypotheses, the term hypothetical may be appliec
to all these cycles. It is desirable to compare the actual pressurevolume
diagram, taken from the indicator card of a steam engine, and the hypothetica
diagram most nearly corresponding with the conditions.
In Fig. 106 is shown in full lines a pressurevolume diagram which ha
WORK OF PISTON ENGINES
367
jeen produced from an actual indicator card taken from a simple noncon
lensing, fourvalve engine having 5 per cent clearance.
Finding the highest pressure on the admission line A'B' and the lowest
pressure on the exhaust line D'E', these pressures are regarded as (initial
pressure) and (back pressure) and a hypothetical diagram constructed cor
responding to Cycle III, with cutoff and compression at the same fraction of
stroke as in the actual engine.
The first difference between the hypothetical and actual PV diagrams
is that the point of release C' is not at the end of stroke, as was assumed for the
hypothetical release, C, a difference which is intentional, since it requires time
jfor pressure to fall after release to the exhaust pressure. This same fact may
icause the corner of the diagram to be rounded instead of sharp as at D.
[Similarly, the point of admission F r is before the end of the return stroke has
20
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/
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X
R
1
J
^^^
r~n
s
r\
A
V"
p'
B'\
5
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':
\
1
^

,c
H
v
E'
C^
i
\jn
b
H
i
FIG. 106. Diagram to Illustrate Diagram Factors.
)een reached, and for a similar reason the corner A' may be rounded, though
\ release and admission are made sufficiently early the corners D f and A' will be
harp, approaching the hypothetical, H and A.
i These differences, however, have little effect upon the area of the actual
tiagram, which is seen to be much smaller than the hypothetical. This
jeficiency of area is the net result of a large number of influences, only a few
I which can be fully explained in connection with the pressure volume
fccussion.
[ Beginning with the point of admission, F', the line F'A'B' represents
he period of admission. The rounding at A' has been explained; the indina
Jon of the line from A' toward B' is due in part to the frictional loss of
sure as the fluid passes at high velocity through ports and passages from
am chest to cylinder. As the stroke progresses, the linear velocity of the
n increases toward midstroke, requiring higher velocities in steam
ages. The greater consequent friction causes pressure to fall in the
inder. The resistance of pipes and valves leading to the engine have
368 ENGINEERING THERMODYNAMICS
an effect on the slope of this line. As cutoff is approached, this pressure
fall becomes more rapid, due to the partial closure of the admission valve.
From B', the point of cutoff, to C", the point of release, is the period
of expansion, during which the pressures are much lower than during the
hypothetical expansion line BC, due principally to the lower pressure at the
point of cutoff B' than at B. Hence, the frictional fall in pressure during
admission has a marked effect upon the work done during expansion. The
curve B'C' rarely follows the law PV = const, exactly, though it commonly
gives approximately the same work area. During the first part of expansion,
the actual pressure commonly falls below that indicated by this curve, but
rises to or above it before the expansion is complete. This is largely due
to condensation of steam on the cylinder walls at high pressures, and its
reevaporation at lower pressure, to be studied in connection with a thermal
analysis of the cycle. The curve of expansion may also depart from this
very considerably, due to leakage, either inwardly, through the admission
yalve, or by piston from a region of higher pressure, or outwardly, through
exhaust valve, or by piston into a region of lower pressure, or by drain,
indicator, or relief valves, or by stuffingboxes.
From the opening of the exhaust valve at the point of release, C', till
its closure at compression E f , is the period of exhaust. Pressures during this
period, as during admission, are affected by frictional losses in the passages
for steam, in this case the pressure in the cylinder being greater than that
in exhaust pipe due to friction, by an increasing amount, as the velocity of;
the piston increases toward midstroke. Thus the line DE' rises above the
line DE until the partial closure of the exhaust near the point of compres
sion causes it to rise more rapidly. r ;
At the point of compression E' the exhaust valve is completely closec
and the period of compression continues up to admission at F r . Leakage
condensation, and reevaporation affect this line in much the same way at
they do the expansion, and often to a more marked degree, due to the facij
that the volume in cylinder is smaller during compression than during
expansion, and a given weight condensed, reevaporated, or added or removec
by leakage will cause a greater change in pressure in the small weigh,
present than if the change in weight had occurred to a large body of steam!
In the compound engine all these effects are present in each cylinder iir
greater or less degree. In addition, there are losses of pressure or of volum
in the receivers themselves between cylinders, due to friction or conden
sation, and where especially provided for, reevaporation by means o
reheating receivers. The effect of these changes in receivers is to cause
loss of work between cylinders, and to make the discharge volume of on I
cylinder greater or less than the supply volume of the next, while the?
were assumed to be equal in the hypothetical cases.
The effect of all of these differences between the actual and hypothetic^
diagrams is to make the actual indicated work of the cylinder something les
than that represented by the hypothetical diagram. Since these effects ai
WORK OF PISTON ENGINES 369
not subject to numerical calculations from data ordinarily obtainable, they
are commonly represented by a single coefficient or diagram factor which is
! a ratio, derived from experiment, between the actual work and that indicated
by hypothesis.
It is at once evident that there may be more than one hypothetical
I diagram to which a certain engine performance may be referred as a standard
I of comparison. When the heat analysis of the steam engine is taken up,
a standard for comparison will be found there which is of great use. For
determination of probable mean effective pressure, however, no method of
calculation has been devised which gives better results than the computa
tion of the hypothetical mean effective pressure from one of the standard
hypothetical diagrams, and multiplying this by a diagram factor obtained
by experiment from a similar engine, under as nearly the same conditions
as can be obtained.
Such diagram factors are frequently tabulated in reference books on the
steam engine, giving values for the factor for various types and sizes, under
various conditions of running. Unfortunately, however, the exact standard
to which these are referred is not stated. In this text it will be assumed,
unless otherwise stated, that the diagram factor for an actual engine is the
ratio of the mean effective pressure of the actual engine to that computed
for Cycle I, without clearance or compression, logarithmic law, with cutoff
at the same fraction of stroke as usual, initial pressure equal to maximum
during admission in actual, and back pressure equal to minimum during exhaust
of the actual engine.
This is selected as the most convenient standard of comparison for mean
effective pressures, as it is frequently impossible to ascertain the clearance
in cases where data are supplied. When it is possible to do so, however, closer
approximation may be made to the probable performance by comparing
Ihe actual with that hypothetical diagram most nearly approaching the cycle,
Using same clearance, cutoff, and compression as are found in the actual.
Commercial cutoff is a term frequently used to refer to the ratio of the
volume AK to the displacement, Fig. 106, in which the point K is found on
E initial pressure line AB, by extending upward from the true point of cutoff
B' a curve PV = const.
While the diagram factor represents the ratio of indicated horsepower
to hypothetical, the output of power at the shaft or pulley of engine is less
Irian that indicated in the cylinders, by that amount necessary to overcome
Mechanical friction among engine parts. If this power output at shaft or
pulley of engine is termed brake horsepower (B.H.P) then the ratio of this to
indicated horsepower is called the mechanical efficiency, E m , of the engine
(583)
The difference between indicated and shaft horsepower is the power
imed by friction (F.H.P.). Friction under running conditions consists
370
ENGINEERING THERMODYNAMICS
of two parts, one proportional to load, and the other constant and inde
pendent of load, or
(F.H.P.) =N[(const.) x(m.e.p.) + (const.)2],
where N is speed, revolutions per minute. But NX (const.) (m.e.p.) = (I.H.P.)^i
and
(F.H.P.) = (I.H.P.)^i+A r (const.) 2 , (584)
where KI and (const.)2 are constants to be determined for the engine, whose
values will change as the conditions of the engine bearingsurfaces or lubri
cation alters. This value for (F.H.P.) may be used to evaluate E m ,
_(I.H.P.) (F.H.P.) _ _ JV(const.) 2
~ I.H.P. Al I.H.P. ' ' ' *
(585)
but indicated horsepower divided by speed is proportional to mean effective
pressure, so that
K 2
~7~~ \
(m.e.p.)
(586)
LOO
.90
.80

"
\
 30
02 1 6 8 10 12 U 16 18 20 22 24 26 28 30 32 31 36 38 10
Mean Effective Pressure
FIG. 107. Diagram to Show Relation of Mechanical Efficiency and Mean Effective Pressure.
From this expression, speed has been eliminated, which agrees with general
observation, that mechanical efficiency does not vary materially with speed.
Values of the constants K\ and K% may be ascertained if (m.e.p.) and E m
are known for two reliable tests covering a sufficient range, by inserting their
values forming two simultaneous equations.
The numerical values of KI found in common practice are between .02
and .05, and for Kz between 1 .3 and 2, in some cases passing out of this range
In Fig. 107 is shown the form of mechanical efficiency curve when plotted
on (m.e.p.) as abscissas, using Ki = M, X 2 = 1.6. It may be noted that a1
WORK OF PISTON ENGINES 371
ter (m.e.p.) the curve does not approach unity, but the value (1 K\)
a limit. The mechanical efficiency becomes zero for this case, at a mean
ective pressure of about 1.67 Ibs. per square inch, which is that just
j sufficient to keep the engine running under no load. For a given speed and size
)f cylinders, the abscissas may be converted into a scale of indicated
jiorsepower.
Prob. 1. Assuming a back pressure of 10 Ibs. per square inch absolute, a clearance
; jf 8 per cent, a cutoff of 40 per cent, and compression of 20 per cent, show how
'm.e.p.) varies with initial pressure over a range of 200 Ibs., starting at 25 Ibs.
Prob. 2. For an initial pressure of 150 Ibs. per square inch absolute, show how
m.e.p.) varies with back pressure over a range of 30 Ibs., starting at Ib. per square
toch absolute, keeping other quantities as in Prob. 1.
Prob. 3. For values of initial pressure, back pressure, etc., as given in Probs. 1
ind 2, show how (m.e.p.) varies with clearance from 1 per cent to 15 per cent.
Prob. 4. For values of initial pressure, etc., as given in Probs. 1 arid 2, show how
pi.e.p.) will vary with cutoff from to 1.
Prob. 5. For values of initial pressure, etc., as given in Probs. 1 and 2, show how
'm.e.p.) will vary with compression for values from to 50 per cent.
Prob. 6. A certain engine developing 675 I.H.P. at a speed of 151 R.P.M.,
[delivered at the shaft 606 H.P. measured by an absorption dynamometer. A second
Wtet'at 100 R.P.M. gave 150 I.H.P., and 114 shaft H.P. If this engine is to deliver
[500 H.P. at the shaft at a speed of 150 R.P.M., what will be the I. H.P. and the mechan
jical efficiency?
Prob. 7. A compound Corliss engine, 25 and 52 ins. diameters, 60 ins. stroke,
doubleacting, was designed for 650 I.H P. at 63 R.P.M. It was found that at this
Aeed and I.H.P. the mechanical efficiency was 91 per cent. When running with no
load, the cylinders indicated 38.1 I.H.P. at 65 R.P.M. Find the probable mechanical
ffrciency when developing 300 I.H.P. at a speed of 64 R.P.M.
24. Consumption of the Steam Engine and its Variation with Valve Move
ment and Initial Pressure. Best Cutoff as Affected by Condensation and
leakage. The weight of steam used by a steam engine per hour divided by
Ihe indicated horsepower is said to be the water rate or steam consumption
of that engine. It is almost needless to say that this is not a constant for
a given engine, since it will change with any change of initial pressure, back
ftessure, or valve setting, leakage, or temperature conditions. Since there
ye at least two other uses of terms water rate or consumption, this may be
ftrmed the actual water rate, or actual consumption, the latter being a more
general term which may refer to the weight of fluid used per hour per
licated horsepower, whatever the fluid may be, steam, air, carbon dioxide,
any other expansive fluid. The present discussion has special reference
steam.
From the hypothetical diagram, by computations such as are described
br the various foregoing cycles, may be obtained a quantity representing
pe weight of fluid required to develop one horsepower for one hour, by the
372 ENGINEERING THERMODYNAMICS
performance of the hypothetical cycle. This may be termed the hypothetico
consumption or for steam cycles the hypothetical water rate.
By the use of the actual indicator card, may be obtained, by methods sti]
to be described, the weight of fluid accounted for by volumes and pressure
known to exist in the cylinder, this being called the indicated consumption o
the engine or indicated water rate if the fluid be steam.
The heat analysis of the steamengine cycle will lead to another standan
of comparison which is of the greatest importance as a basis of determining
how nearly the actual performance approaches the best that could b<
obtained if the engine were to use all available energy possessed by the steam
At present the object is to compare the actual and indicated performance wit!
that hypothetically possible with cylinders of the known size. Accordingly
attention will be confined first to hypothetical consumption, and the quan title;
upon which it is dependent.
For Cycle III, which is the most general for the singleexpansion engine
logarithmic law, the expression for consumption in pounds fluid per hour pei
indicated horsepower, found in Section 5, Eq. (267), is as follows:
Hypothetical consumption, Ibs. per hr. per I . H.P.
in which the value of mean effective pressure itself depends upon (in.pr.)
(bk.pr.), c, Z, and X. The density of the fluid at initial pressure, d lf is to b
ascertained from tables of the properties of steam or of whatever fluid is used.
In Fig. 108 are the results of computations on the hypothetical steam con
sumption, using mean effective pressures as plotted in Fig. 105. For each curve
conditions are assumed to be as stated on the face of the diagram, varying onl
one of the factors at a time.
Other conditions remaining unchanged, it may be noted that consumptio
decreases for an increase of initial pressure, though not rapidly in the highe
pressure range.
Cutoff has a marked effect upon consumption, the minimum occurring
when cutoff is such as to give complete expansion. This occurs when
1+c ___ (in.pr.)
Z'+c~ (bk.pr.)'
or
(in.pr.)
which may be termed hypothetically best cutoff. In the case assumed in til
diagram,
WORK OF PISTON ENGINES
373
If clearance be varied, maintaining constant compression and cutoff, large
trance will give high consumption due to an excessive quantity of fluid
equired to fill the clearance space. Extremely small clearance leads to a high
pressure at the end of compression, causing a loss of meaneffective pressure,
ijid consequent high consumption. Between, the consumption has a minimum
>oint, which is dependent for its location on both cutoff and compression.
Decreasing back pressure has a beneficial effect upon mean effective pressure
,md consumption. This would be still more marked in the figure if a case had
>een selected with a very short cutoff.
Compression, throughout the ordinary range of practice, has but slight effect
ipon consumption, indicated by the flat middle portion of the curve in Fig.
35
30
25
20
15
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INITIAL PRESSURE 100 LBS.ABS
BACK PRESSURE 15
CLEARANCE 10
CUT OFF .25
COMPRESSION .35
:
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5 10 15 20 25 30 .1 2 .3 .4 .5 .6 .7 .8 .9 1.0
ho. 108. Curves to Show the Variation of Hypothetical Steam Consumption of Simple
Engines, Logarithmic Expansion and Compression.
08. Very small or zero compression permits too much highpressure steam to
K admitted to the clearance space without doing work, and excessively large
learance causes pressures during compression to rise very high, thereby de
Teasing mean effective pressure; hence this curve of consumption rises at
poth ends.
Hypothetically, the best attainable consumption for given initial and back
jessures is obtained when both expansion and compression are complete.
I The indicated consumption, or, as it is frequently called for the steam engine,
fsteam accounted for by the indicator card" " or " indicated water rate," is
itermined from the indicator card as follows. Let Fig. 109 represent an indicator
flagram. The points of cutof #nd compression are located from the form of
374
ENGINEERING THERMODYNAMICS
the line, at the highest point on the expansion line and the lowest point on th'
compression line respectively. The fraction of the card lengths completed a
cutoff,
AD>
and the fraction of card length from point of compression to end of stroke,
are determined, the pressure at cutoff and compression measured by th
proper vertical scale, and the corresponding densities, BI, and 2 respectively
are ascertained from steam tables for dry saturated steam. Clearance, Cl, i
known or ascertained by the form of the compression curve (Chap. I, Section 12
o
CutOff
lAtm.
B C
D V
FIG. 109. Diagram to Illustrate Method of Determining Indicated Water Rate of Sten
Engine.
>
At the point of cutoff, the weight of dry saturated steam present in t
cylinder is D(Z+c)Bi, and at compression the weight present is D(X+c)
on the assumption that the steam in the cylinder is of density 81 and 82
these two instants. Accepting this assumption, the weight of steam used j:
cycle is
Wt. steam per cycle
The work per cycle
[(Z+c)$i (X+c)Z 2 ]D^
TF=144D(m.e.p.),
and for n cycles per minute the indicated horsepower is
T IT P _144n/Xm.e.p.)
i.n.r  .>., ()()n
375
_wm_ = 60X33,OOOXD[(Z+c)Bi(X+c)o 2 ]n
I. H. P. 144wD(m.e.p.)
or,
Ind. consumption, Ibs. per hr. per I.H.P.
(589 >
which is the expression used to find indicated consumption for either simple
or multipleexpansion engines. In applying this to the multipleexpansion
engine the terms Z, X and c are found for any one cylinder, and the mean effective
pressure is referred to that cylinder. There may be, therefore, as many computa
tions as there are expansion stages. For a compound engine, for instance,
indicated consumption according to highpressure card is found by inserting
in formula Z, X and c for the highpressure card, 81 and 82 for corresponding
pressures, and for (m.e.p.) use
(m.e.p. ref. to H.P.) = (m.e.p.)^+(m.e.p.) L ^. . . . (590)
If the computation is done by means of events on the lowpressure card,
the (m.e.p.) must be referred to the low.
(m.e.p. ref. to L.P.) = (m.e.p.)/r^+(m.e.p.)z, (591)
In general for a multipleexpansion engine
(m.e.p. ref. to cyl. A) = 2(m.e.p.)yc (592)
jt//
It is often difficult and sometimes impossible to determine the point of
i cutoff and of compression on the indicator card. The expansion and compres
sion lines are of very nearly hyperbolic in form and are usually recognizable.
The highest point on the hyperbolic portion of the expansion line is regarded as
cutoff, and the lowest point on the hyperbolic portion of the compression line,
as the point of compression. It must be understood that by reason of the
[condensation and reevaporation of steam in cylinders the weight of steam
>roper is not constant throughout the stroke, so that calculations like the
ibove will give different values for every different pair of points chosen. The
lost correct results are obtained when steam is just dry and these points are
it release and compression most nearly.
376 ENGINEERING THERMODYNAMICS
When under test of actual engines the steam used is condensed and weighed
and the indicated horsepower determined, then the actual steam consumption
or water rate can be found by dividing the weight of water used per hour in
the form of steam by the indicated horsepower. This actual water rate is
always greater than the water rate computed from the equation for indicated
consumption. The reasons for the difference have been traced to (a) leakage
in the engine, whereby steam weighed has not performed its share of work, to
(b) initial condensation, whereby steam supplied became water before it could
do any work, (c) variations in the water content of the steam by evaporation
or condensation during the cycle, whereby the expansion and compression laws
vary in unpredictable ways, affecting the work.
Estimation of probable water rate or steam consumption of engines cannot,
therefore, be made with precision except for engines similar to those which have
been tested, in all the essential factors, including, of course, their condition,
and for which the deficiencies between actual and indicated consumptions
have been determined. This difference is termed the missing water, and end
less values for it have been found by experiments, but no value is of any use
except when it is found as a function of the essential variable conditions that
cause it. No one has as yet found these variables which fix the form for an
empiric formula for missing water nor the constants which would make such a
formula useful, though some earnest attempts have been made. This is nc
criticism of the students of the problem, but proof of its elusive nature, and the
reason is probably to be found in the utter impossibility of expressing by
formula the leakage of an engine in unknown condition, or the effect of its
condition and local situation on involuntary steam condensation and evaporai
tion. It is well, however, to review some of these attempts to evaluate J
missing water so that steam consumption of engines may be estimated
After studying the many tests, especially those of Willans, Perry announcer
the following for noncondensing engines, in which the expansion is but littlt
Missing water Z
Indicated steam J  / ~^ T) '
where d is the diameter of the cylinder in inches and TV the number of revolution
per minute. This indicates that the missing steam or missing water has bee
found to increase with the amount of expansion and decrease with diameter c
cylinder and the speed. Thermal and leakage conditions are met by the us
of difference values of m, for there are given
m = 5 for welljacketed, welldrained cylinders of good construction wit
four poppet valves, that is, with minimum leakage and condensatior
w = 30 or more for badly drained un jacketed engines with slid, valves,
is, with high leakage and condensation possibilities.
m = 15 in average cases.
WORK OF PISTON ENGINES 377
For condensing engines Perry introduces another variable the initial pres
ij sure pounds per square inch absolute, p giving
Missing water = __ _ _
Indicated steam
It might seem as if such rules as these were useless, but they are not, especially
when a given engine or line of engines is being studied or two different engines
I compared; in such cases actual conditions are being analyzed rather than
j predictions made, and the analysis will always permit later prediction of con
siderable exactness, if the constants are fixed in a formula of the right empiric
form. Similar study by Heck has resulted in a different formula involving
different variables and constants, but all on the assumption that the dis
crepancies are due to initial condensation. He proposes an expression equi
valent to
Missing steam _ .27 IS(x2xi)
Indicated steam ~~ ^/]y\ piZ '
in which ]V = R.P.M. of the engine;
d = diameter in inches;
L = stroke in feet;
S = the ratio of cylinder displacement surface in square feet to dis
placement in cubic feet.
L d '
The term (x 2 Xi) is a constant supposed to take into account the amount
Srf initial condensation dependable on the difference between cylinder wall
knd livesteam temperature and is to be taken from a table found by trial as
Ihe difference between the x for the high pressure and x for the low pressure,
both absolute, see Table XIV at the end of the Chapter.
In discussing the hypothetical diagrams, it was found that best economy was
[.obtained with a cutoff which gives complete expansion. For other than
hypothetical diagrams this is not true, which may be explained most easily by
inference to the curves of indicated, and actual consumption, and missing
(jBteam, Fig. 110.
The curve ABC is the hypothetical consumption or water rate for a certain
steam engine. Its point of best economy occurs at such a cutoff, B, that expan
378
ENGINEERING THERMODYNAMICS
sion is complete. The curve GUI is computed by Heck's formula for missing
water. The curve falls off for greater cutoffs. Adding ordinates of these tw<
curves, the curve DEF for probable consumption is found. The minimun
point in this curve, E, corresponds to a longer cutoff than that of ABC
Since cutoff B gave complete expansion, cutoff E must give incomplete expan
sion. In other words, due to missing steam, the condition which really give
least steam consumption per hour per indicated horsepower corresponds to j
release pressure, which is greater than the back pressure.
It should be noted that the minimum point mentioned above will not b<
best cutoff, for the output of the engine is not indicated, but brake horsepower
30
1
S80
1
10
}
\
A
N
\v
I
.
. '
==
. =
C
\
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\
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1
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Per Cent Cut off
FIG. 110. Diagram to Show Displacement of Best Cutoff Due to Effect of Missing Wate
from Point B for the Hypothetical Cycle to Some Greater Value E.
In Fig. Ill on cutoff as abscissa are plotted (EFG) consumption pound
per hour per I.H.P., and for the case assumed, (OD) the curve of mechanic;
efficiency, based on cutoff.
(Ibs. steam per hr.) . B.H.P. _ (Ibs. steam per hr.)
I.H.P. *I.H.P.~ B.H.R '
or, in other words,
Consumption, Ibs. per hr. per I.H.P.
  = Consumption, Ibs. per hr. per B.H.P. (59('i
Due to the increasing value of E m for greater cutoffs, the minimum point
corresponds to a cutoff still longer than for the curve EFG, which itself W
found hi Fig. 110 to give a longer cutoff than that of the hypothetical curve
Hence the best cutoff for economy of steam, where the net power at tl
shaft is regarded as the output, will be such as to give incomplete expansion
or a release pressure above back pressure, this effect being caused by bot
missing steam and by frictional losses.
Prediction of actual consumption of steam engines as a general propositk:
is almost hopeless if any degree of accuracy worth while is desired, though tl
WORK OF PISTON ENGINES
379
effect on steam consumption of changing the value of any one variable can be
pretty well determined by the previous discussion qualitatively, that is, in kind, '
though not quantitatively in amount. Probably the best attempt is that
of Hrabak in German, which takes the form of a large number of tables
developed from actual tests though not for engines of every class. These tables
are quite extensive, being in fact published as a separate book and any abstrac
tion is of no value.
There is, however, a sort of case of steam consumption prediction that can be
carried out with surprising precision and that is for the series of sizes or line
of engines manufactured by one establishment all of one class, each with
about the same class of workmanship and degree of fit, and hence having
leakage and cylinder condensation characteristics that vary consistently through
out the whole range. For such as these tables and curves of missing water
o .1 .2 .3
Per Cent Cut off
FIG. 111. Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cutoff.
can be made up and by the best builders are, for making guarantees of steam
consumption for any service conditions that their engines are able to meet.
The practice of one firm making what is probably the best line of stationary
engines in this country is of sufficient interest to warrant description. The
primary data are curves of indicated water rate plotted to mean effective
pressure for clearances of three or four per cent, and that mean effective pressure
is chosen in any one specific case that will give the horsepower desired at the
fixed speed for some one set of cylinder sizes avaliable. To this indicated
water rate a quantity is added constituting the missing water which is made
up of several parts as follows: The first is an addition representing condensa
tion which is plotted in curve form as a function of (a) boiler pressure, (6)
superheat in the steam, (c) piston speed, (d) the class of engine simple, com
pound or triple, with jacketed or unjacketed cylinders, and for cylinder ratios
from 4 to 1, to 6 to 1 in the case of compounds. It is therefore a most complex
quantity, the nature of the variations in which can only be indicated here.
380 ENGINEERING THERMODYNAMICS
For example, increase of piston speed decreases the condensation loss as
does multiple expansion, and also jacketing, while increase of superheat in the
steam also decreases it, but superheat has less effect in triple than in com
pounds and less in compound than in simple engines.
The next factor of correction is that covering leakage losses, also additive
to indicated water rate and which with it and the condensation loss make
up the probable steam consumption. The leakage decreases regularly with
increase of piston speed, is less for large than for small engines, the change
being rather fast from 50 to 200 horsepower and much slower later, being
scarcely anything at all over 2000 horsepower.
Example 1. What cutoff will give the lowest indicated water rate for a 9x12
in. engine, with 5 per cent clearance and no compression when running noncondensing
on an initial pressure of 100 Ibs. per square inch gage, and what will be the value
of this water rate? What steam will be used per hour per brake horsepower
hypothetically? From Eq. (587)
= (1+.05) .05=8.7 per cent,
115
and
(m.e.p.) =115.087X(.087+.05) log^fl 15=27.2 Ibs. sq.in.
L .lo7J
Hence
Steam per hour per I.H.P. = ~  .137. 05 X^ X. 262 = 17.2 Ibs.
2il ,2i [ lloj
From the curve of Fig. 107, assuming it to apply to the engine, for this value of
(m.e.p.) mechanical efficiency is 90 per cent, hence from Eq. (596) the weight of
steam per shaft horsepower per hour will be 19.1 pounds.
Prob. 1. Draw diagram similar to Fig. 108 for following case:
Initial pressure, 135 Ibs. per square inch gage, back pressure 10 Ibs. per square
inch absolute, clearance 5 per cent, cutoff 30 per cent, compression 25 per cent.
Prob. 2. From indicator diagram shown in Fig. 106 find the indicated water rate
of the engine from which it was taken.
Prob. 3. The indicated water rate of a 9xl2in. jacketed engine when running
noncondensing at a speed of 250 R.P.M. with an initial pressure of 100 Ibs. per
square inch gage and \ cutoff is 50 Ibs. Using Perry's formula what will be the
probable actual steam used by engine per horsepower hour "
Prob. 4. A 24 x48in. engine in good condition is found to have an indicated water
rate of 25 Ibs. when cutoff is i, initial pressure 100 Ibs. per square inch gage, back
pressure 10 Ibs. per square inch absolute, and speed of 125 R.P.M. What will be the
missing water, and the rate as found by Perry's formula and by Heck's?
WORK OF PISTON ENGINES 381
Prob. 5. What will be the probable amount of steam used per hour by a 36x48
in. engine with 5 per cent clearance running at 100 R.P.M. on an initial pressure of
150 Ibs. per square inch gage a back pressure of 5 Ibs. per square inch absolute, \ cut
off and 10 per cent compression?
Prob. 6. How will the amount of steam of Prob. 5 compare with that used by
a 15X22x36in engine with 5 per cent clearance in each cylinder, running at 100
R.P.M. on same pressure range with J cutoff in highpressure cylinder, \ cutoff in
low, and 10 per cent compression in each cylinder?
25. Variation of Steam Consumption with Engine Load. The Willans
Line. Most Economical Load for More than One Engine and Best Load
Division. However valuable it may be to the user of steam engines to have
an engine that is extremely economical at its best load which, it should be
noted, may have any relation to its rated, horsepower, it is more important
usually that the form of the economy load curve should be as flat as possible
and always is this case when the engine must operate under a wide range of
load. This being the case it is important to examine the real performance
curves of some typical engines all of which have certain characteristic
similarities as well as differences.
From the discussion of hypothetical and indicated water rates it appears
that the curve of steam consumption (vertical) to engine load (horizontal)
is always concave upward and always has a minimum point, not at the maxi
 mum load. Actual consumption curves are similar in general form, but as
has been pointed out, the load at which the water rate is least corresponds
to some greater mean effective pressure than that for the hypothetical, so
the whole curve is displaced upwards and to the right by reason of cylinder
condensation and leakage losses. This displacement may be so great as to
prevent the curve rising again beyond the minimum point, in which case the
least steam consumption corresponds to the greatest load. Just what form
(the actual water rateload curve will take depends largely on the form of valve
gear and type of governing method in use, by throttling initial pressure with
a fixed cutoff or, by varying cutoff without changing initial pressure, with or
without corresponding changes in the other valve periods.
Whenever the control of power is by throttling of the supply steam the
curve is found to be almost exactly an hyperbola, so that (water rateXhorse
Jpower) plotted to horsepower is a straight line which being characteristic
 is much used in practical work and is known as the Willans line. All other
 engines, that is, those that govern on the cutoff, have Willans lines that
j are nearly straight, such curvature as exists being expressed by a second
i degree equation instead of one of the first degree.
Equations for Willans lines can always be found for the working range
jtof load, that is, from about half to full load, though not for the entire range,
fccept in unusual cases, and these equations are of very great value in pre
dicting the best division of load between units, which is a fundamental step
[In deciding, how many and what sizes of engine to use in carrying a given
load in industrial power plants.
382
ENGINEERING THERMODYNAMICS
\\
\
m
o o
5* o5
'd *H 'I Joel rawo^s jo '{
WORK OF PISTON ENGINES
3S3
Before taking up the derivation of equations some actual test curves will
>e examined and a number of these are grouped in Fig. 112 for engines of
iarious sizes, simple and compound, up to 10,000 H.P., on which vertical
distances represent pounds of steam per hour, per I.H.P. and horizontal
.H.P. To show the essential similarity of the curves for engines of different
ize more clearly, these are replotted in Fig. 113 to a new load scale based
;>n best load of each, which is taken as unity. This is evidently a function of
{lean effective pressure, just what sort of function does not matter here. In
75 loo 125
Percentage of most economical load
150
IG. 113. Typical Water Rate Lead Curves for Steam Engines Plotted to Fractional Loads.
ery case the Willans line is also plotted in Fig. 112, each line being num
3red to correspond to its water rate curve.
[ As there is a corresponding similarity of form for the water rate and
fcllans line of steam turbines, though the reasons for it will be developed
fcr, it must be understood that the mathematical analysis that follows applies
both turbine and piston steam engines, and finally it makes no difference
jiat units are used for load, whether I.H.P., or B.H.P. or K.W. of a direct
mnectecl electric generator.
In Fig. 114 is shown the waterrate curve to a K.W. base for the 10,000 K.W.
Irtis steam turbine at the Chicago Edison, Fiske Street Station for which the
llowing equation fits exactly:
384
ENGINEERING THERMODYNAMICS
where Y = pounds of steam per hour r 1000;
P = load (in this case in K.W.H1000;
Y
p = pounds of steam per K.W. hr.
A similar equation fits fairly well the curve of Fig. 115, representing tl
7000 H.P. piston engines of the Interborough Railway, Fiftyninth Strei
station, as well as the combined piston engine and lowpressure steam turbii
400
4oOO 5000
7000 9000 11000
1000 P 5= Kilowatts
13000
15000
FIG. 114. Performance of a 10000K.W. Steam Turbine.
taking its exhaust steam, in the same station, but with different numeric
constants, as below:
Y 36 7^
Piston engine, p = ^
Combined piston engine and turbine, B = ~T>'  2.90+.713P.
A third case of smaller size is shown in Fig. 116, representing the p
formance of a 1000K.W. Corliss piston engine driving a generator for wh
the equation is
WORK OF PISTON ENGINES
Kilowatts
385
7000 8000 9000 10000 11000 12000 13000 14000 15000 16000
Kilowatts =1000 P
FIG. 115. Performance of a 7000H.P. Piston Engine alone and with a Lowpressure
Steam Turbine.
/
\
/
/
s
/
25 24
^ /
\
,>,
/
II
\
>
/
^l 
\
g
A
V
\
g
>'/
20 ^20
e
\
\
*>/
&
s *
\\
/
o ^
\
t
/
/
v, o3
* 3
\
/
/
^ ij
\
/
S
S 15 ^ 18
g
X
^
,s>
^
^ ^
"3
X
^^
*^^
^
^*^
1
X
^
9
10 16
JOG
K)
a
K)
i;
Kilowatts = 1000 P
FIG. 116. Performance of a 1000K.W. Steam Turbine.
386 ENGINEERING THERMODYNAMICS
These illustrations could be multiplied indefinitely, but those given will
suffice to establish the fact that the two following equations are fundamental
over the working range of any steam engine of whatever type:
Water rate line, j=~+B+CP . ........ (597)
Water per hour, Willans line, Y = A+BP+CP 2 , . . (598
in which F is the weight of steam per hour and P the engine load whether
expressed in indicated or brake horsepower, or in kilowatts.
At the most economical load the water rate is a minimum, so that
A
dP
whence the most economical load is
U
(599)
Where the Willans line is straight, C = 0, and the most economical load
is the greatest load.
Two engines carrying the same load must divide it and some one pro
portion may be best. To find out, consider first any number of similar engines,
that is, engines that have the same constants A, B, and C, denoting each case
by subscripts. Then
Let P = total load;
" PI, P 2 , P 3 , etc. = individual engine loads;
F = total water per hour;
" YI, F 2 , F 3 = water per hour for each engine.
Then
.+ F n
. Pn) + C(F
P 3 2 + . . .+P n 2 ).
Only the last term is variable and this is a minimum when
Therefore for similar engines, the best division of load is an equal division.
WORK OF PISTON ENGINES 387
When the engines are dissimilar it is convenient to first consider the case
straight Willans lines for which C = 0. Then for two such engines
Y = A 1 +A 2 +B 1 P 1 +B 2 P 2
= (A l +A 2 )+B l (PP 2 )+B 2 P 2
= (A l +A 2 )+B 1 P+(B 2 B 1 )P 2 .
t any given load P the first two terms together will be constant, and the
r per hour will be least when the last term is least. As neither factor
i be zero, this will occur when P 2 is least.
Therefore for two dissimilar engines the best division of load is that which
ts the greatest possible share on the one with the smaller value of B, in its equation,
yvided each has a straight Willans line.
Two dissimilar engines of whatever characteristics yield the equation,
Y = A 1 +A 2 +B l P l +B 2 P 2 +CiP i 2 +C 2 P 2 2
= (Ai+A 2 +B ] P+C 1 P 2 )
+ (B 2 2PC 1 B l )P 2
+(Ci+C 2 )P 2 2 .
Differentiation with respect to P 2 , and solving for P 2 , the load for the second
gine that makes the whole steam consumption least, gives,
(600)
= constant + constant X P.
Therefore, the load division must be linear and Eq. (600) gives the numerical
ue, when any two engines share a given load.
This sort of analysis can be carried much further by those interested, but
tee forbids any extension here. It is proper to point out, however, that
means of it the proper switchin points for each unit in a large power station
L be accurately found, to give most economical operation on an increasing
tion load.
i 26. Graphical Solution of Problems on Horsepower and Cylinder Sizes .
je diagram for mean effective pressures in terms of initial and back
(assure, clearance, compression and cutoff, Fig. 117, facilitates the solution
1 3q. (262) in Section 5. The mean effective pressure is the difference
\reen mean forward and mean back pressure. The former is dependent
i clearance, cutoff and initial pressure. In the example shown on the
fe by letters and dotted lines, clearance is assumed 5 per cent, shown at
Project horizontally to the point F, on the contour line for the assumed
pff, 12 per cent. Project downward to the logarithmic scale for " mean
388
ENGINEERING THERMODYNAMICS
'I
o
>
s
i
I
Q
1
^3
WORK OF PISTON ENGINES 389
ard pressure in terms of initial pressure " to the point G. On the scale
" initial pressure " find the point H, representing the assumed initial pres
ure, 115 Ibs. absolute. Through G and H a straight line is passed to the point
C on the scale for " mean forward pressure/' where the value is read,
i.f.p. = 49.5 Ibs. absolute.
Mean back pressure is similarly dependent upon clearance, compression
iid back pressure, and the same process is followed out by the points A, B,
1, D and E, reading the mean back pressure, 3.2 Ibs. absolute at the point E.
?hen by subtraction,
(m.e.p.) = (m.f. p.) (m.b.p.) =49.53.2 = 46.3 Ibs.
Fig. 118 is arranged to show what conditions must be fulfilled in order to
f<btain equal work with complete expansion in both cylinders in a compound
ngine, finite receiver, logarithmic law, no clearance, Cycle VII, when low
ressure admission and highpressure exhaust are not simultaneous. This is
ecussed in Section 11, and the diagram represents graphically the conditions
xpressed in Eqs. (376), (377), (378), (379).
To illustrate its use assume that in an engine operating on such a cycle,
le volume of receiver is 1.5 times the highpressure displacement, 1.5 = 2/, then
=.667. Locate the point A on the scale at bottom of Fig. 118, corresponding
o this value. Project upward to the curve marked "ratio of cutoffs " and at
le side, C, read ratio of cutoffs
= .572.
text extending the line AB to its intersection D, with the curve GH, the point
is found. From D project horizontally to the contour line representing the
iven ratio of initial to back pressure. In this case, initial pressure is assumed
in times back pressure. Thus the point E is located. Directly above E at
ie top of the sheet is read the cylinder ratio, at F,
If cylinder ratio and initial and final pressures are the fundamental data of
problem, the ratio of cutoffs and ratio of highpressure displacements
receiver volume may be found by reversing the order.
390
ENGINEERING THERMODYNAMICS
WORK OF PISTON ENGINES 391
GENERAL PROBLEMS ON CHAPTER III.
Prob. 1. How much steam will be required to run a 14xl8in. doubleacting
engine with no clearance at a speed of 200 R.P.M. when initial pressure is 100 Ibs.
per square in. gage, back pressure 28 ins. of mercury (barometer reading 30 ins ), and
cutoff is ? What will be horsepower under these conditions?
NOTE: 5 for 100 Ibs. =.26, for 28 ins. =.0029.
Prob. 2. Draw the indicator cards and combined diagram for a compound steam
engine without receiver, and with 3 per cent clearance in low pressure and 5 per cent in
high, when initial pressure is 100 Ibs. per square inch absolute, back pressure 10 Ibs.
per square inch absolute, highpressure cutoff , highpressure compression yV, and
low pressure compression 5.
Prob. 3. A simple doubleacting engine, 18x24 ins., is running at 100 R.P.M.
on compressed air, the gage pressure of which is 80 Ibs. The exhaust is to atmosphere.
If the clearance is 6 per cent and cutoff f , and compression 10 per cent, what horse
power is being developed, the expansion being adiabatic, and how long can engine be
run at rated load on 1000 cu.ft. of the compressed air?
Prob. 4. Will the work be equally distributed in a 12xl8x24in. engine with
infinite receiver and no clearance when cutoff is  in high pressure cylinder, and f in
low, expansion being logarithmic, initial pressure 150 Ibs. per square inch absolute
and back pressure atmosphere? What will be work in each cylinder?
Prob. 5. The receiver of a 15X20x22 in. engine is 4 times as large as high
pressure cylinder. What will be the horsepower, steam used per hour, and variation
in receiver pressure for this engine, if clearance be considered, zero and initial pressure
is 125 jbs. per square inch gage, back pressure 5 Ibs. per square inch absolute, cutoffs
and f in high and lowpressure cylinders respectively, and piston speed is 550 ft. per
minute?
NOTE: 8 for 125 Ibs. =.315, for 5 Ibs. =.014.
Prob. 6. At no load an engine having 7 per cent clearance, cuts off at 4 per cent
of its stroke, while at full load it cuts off at 65 per cent of its stroke. At no load,
compression is 40 per cent and at full load 5 per cent. What percentage of fullload
horsepower is required to overcome friction, and what percentage of steam used at
full load, is used on friction load, if initial pressure is constant at 100 Ibs. per square
inch gage, back pressure constant at 5 Ibs. per square inch absolute, and expansion
is logarithmic?
NOTE: 8 for 100 Ibs. =.262, for 5 Ibs. =.014.
Prob. 7. The initial pressure on which engine is to run is 115 Ibs. per square inch
gage, and steam is superheated and known to give a value of s = 1.3. For an engine
in which clearance may be neglected, work is to be equal, and expansion complete
in both cylinders, when back pressure is 10 Ibs. per square inch absolute. What must
be the cutoffs and cylinder ratio to accomplish this when receiver is 3 times high
pressure cylinder volume?
Prob. 8. A 12in. and 18x24 ins. doubleacting engine with zero clearance and
infinite receiver operates on an initial pressure of 150 Ibs. per square inch gage, and
392 ENGINEEEING THERMODYNAMICS
a back pressure of 5 Ibs. per square inch absolute. What will be the release and receiver
pressures, horsepower, and steam consumption when speed is 150 R.P M. ; expansion
logarithmic, and cutoff  in each cylinder?
NOTE: 8 for 150 Ibs. = .367, for 5 Ibs. = .014.
Prob. 9. If a third cylinder 24 ins. in diameter were added to engine of Prob. 8
and cutoff in this made , how would horsepower, steam consumption, receiver and
release pressures change?
Prob. 10. What would have to be size of a single cylinder to give same horsepower
at same revolutions and piston speed as that of engine of Prob. 8 under same conditions
of pressure and cutoff?
Prob. 11. With the higiipressure cutoff at \, and low and intermediate cutoffs
at fV, what will be the horsepower, water rate and receiver pressures of a 30 X 48 X 77 X72
in. engine running at 102 R.P.M. on an initial pressure of 175 Ibs. per square inch gage
and a back pressure of 26 *ins. of mercury (barometer reading 30 ins.), if the
receiver be considered infinite and expansion logarithmic, clearance zero? What change
in intermediate and lowpressure cutoffs would be required to give equal work distribu
tion?
NOTE: 8 for 175 Ibs. =.419, for 26 ins. =.0058,
Prob. 12. If it had been intended to have all the cutoffs of the engine of Prob.
11, equal to ^, what should have been the size of the intermediate and lowpressure
cylinders to give equal work for same pressure range and same highpressure cylinder?
Prob. 13. To attain complete expansion in all cylinders of the engine of Prob. 11,
with the initial and back pressures as there given, what cutoffs would be required
and what receiver pressures would result?
Prob. 14. A compound locomotive has no receiver, the highpressure clearance
is 8 per cent, and lowpressure clearance 5 per cent. The cylinders are 22 and
33 X48 ins., highpressure cutoff f , high and lowpressure compression each 10 per
cent, initial pressure 175 Ibs. per square inch gage, back pressure one atmosphere, and
expansion and compression logarithmic. What will be the horsepower at a speed
of 40 miles per hour, the engine having 7ft. driving wheels? At this speed, how
long will a tank capacity of 45,000 gallons last?
NOTE: 5 for 175 Ibs. =.419, for 15 Ibs. =.038.
Prob. 15. A superheater has been installed on engine of Prob. 14 and expansion
and compression, now follow the law PV S =c, when s = 1.2. What effect will this have
on the horsepower and steam consumption?
Prob. 16. What will be the maximum receiver pressure work done in each cylinder
and total work for a crosscompound engine 36 and 66 X48 ins., running at 100 R.P.M.
on compressed air of 100 Ibs. per square inch gage pressure, exhausting to atmosphere
if the high pressure cutoff is , clearance 6 per cent, compression 20 per" cent, low
pressure cutoff is f, clearance 4 per cent, compression 15 per cent, and receiver volume
is 105 cu.ft.?
Prob. 17. A manufacturer gives the horsepower of a 42x64x60in. engine as
2020, when run at 70 R.P.M. on an initial pressure of 110 Ibs. per square inch gage,
atmospheric back pressure, and .4 cutoff in highpressure cylinder. How does this
value compare with that found on assumption of 5 per cent clearance in high, 4 per
cent in low, and complete expansion and compression is each cylinder?
Prob. 18. A mine hoisting engine is operated on compressed air with a pressure of
150 Ibs. per square inch absolute and exhausts to atmosphere. The cylinder sizes are
26X48X36 ins., and clearance is 5 per cent in each. At the start the highpressure
WORK OF PISTON ENGINES 393
toff is 1 and low pressure $, while normally both cutoffs are . The exhaust
from highpressure cylinder is into a large receiver which may be considered infinite.
] The compression is zero at all times. Considering the exponent of expansion to be
i 1.4, what will be the horsepower under the two conditions of cutoff given, for a speed
of 100 R.P.M.?
Prob. 19. What must be ratio of cylinders in the case of a compound engine with
: infinite receiver, to give equal work distribution complete expansion and com
j pression if the least clearance which may be attained is 5 per cent in the high
pressure cylinder, and 3 per cent in the lowpressure. The engine is to run non
; condensing on an initial pressure of 125 Ibs. per square inch gage, with expansion
exponent equal to 1.3? What must be the cutoffs and compressions to satisfy these
j conditions?
Prob. 20. Assuming 7 per cent clearance in highpressure cylinder and 5 per cent
in low, infinite receiver, and no compression, how will the manufacturer's rating of
2100 H.P. check, for a 36X6x48in, engine running at 85 R.P.M. on an initial
pressure of 110 Ibs. per square inch gage, and a back pressure of 26 ins. vacuum, with
.3 cutoff in high pressure cylinder?
Prob. 21. For a 25X40x36in. engine, with 5 per cent clearance, I cutoff
and 20 per cent compression in each cylinder, what will be horsepower for an initial
pressure of 100 Ibs. per square inch gage, and a back pressure of 17.5 Ibs. per square inch
absolute, with logarithmic expansion and compression?
Prob. 22. What must be the cylinder ratio and cutoff to give complete expansion
ki a noclearance, 14 and 22 x24in. engine with no receiver and logarithmic expansion,
when initial pressure is 100 Ibs. per square inch gage, and back pressure 10 Ibs. per
square inch absolute? What will be the horsepower and steam used for these conditions
at a speed of 150 R.P.M.?
NOTE: 5 for 100 Ibs. =.262, for 10 Ibs. =.026.
Prob. 23. A 24X20x24 in. engine with no receiver or clearance, runs on com
pressed air of 120 Ibs. per square inch gage pressure, and exhausts to atmosphere.
When running at a speed of 125 R.P.M. , with highpressure cutoff \, what horse
power will be developed and how many cubic feet of compressed air per minute will
be required to run the engine, the expansion being adiabatic? Will the work be equally
divided between the two cylinders?
Prob. 24. It is desired to run the above engine as economically as possible. What
change in' cutoff will be required, and will this cause a decrease or increase in horse
power and how much? How will the quantity of air needed be affected?
Prob. 25. A mill operates a crosscompound engine with a receiver 3 times as largo
Iks high pressure cylinder, on an initial pressure of 125 Ibs. per square inch gage, and a
lack pressure of 10 Ibs. per square inch absolute. The engine may be considered as
Beithout clearance, and the expansion as logarithmic. As normally run the cutoff in
[thighpressure cylinder is f and in low, J. It has been found that steam is worth 25
I cents a thousand pounds. What must be charged per horsepower day (10 hours)
to pay for steam if the missing water follows Heck's formula?
NOTE 8 for 125 =.315, for 10 = .026.
Prob. 26. By installing a superheater the value of s in Prob. 25 could be changed
1.3. The cost of st am would then be 30 cents a thousand pounds. From the effect
value of s alone would the installation of the superheater pay?
Prob. 27. When a 26X48x36in. crosscompound engine with a receiver volume
35 cu.ft. and zero clearance, is being operated on steam of 125 Ibs. per square inch
394 ENGINEERING THEEMODYNAMICS
gage initial pressure, and atmospheric exhaust, is the work distribution equal, wht
highpressure cutoff is f and lowpressure cutoff ? For these cutoffs what
fluctuation in receiver pressure and what steam will be used per horsepower hour?
NOTE: 8 for 125 = .315, for 15 lbs.=.038.
Prob. 28. To operate engine of Prob. 27 under most economical conditions, wh;
values must be given to the cutoffs, and what values will result for receiver pressure
horsepower, and .steam used per hour?
Prob. 29. What will be the horsepower and steam used by a 20x30x364
engine with infinite receiver and no clearance, if expansion be such, that s = 1.2
highpressure cutoff f ,. lowpressure cutoff , initial pressure 100 Ibs. per square inc
gage, back pressure 3 Ibs. per square inch absolute, and speed 100 R.P.M.
NOTE: S for 100 Ibs. =.262, for 3 Ibs. = .0085.
Prob. 30. The following engine with infinite receiver and noclearance, runs c
steam which expands according to the logarithmic law. Cylinders 9, and 13 x:
ins., initial pressure 125 Ibs. per square inch gage, back pressure 5 Ibs. per square in<
absolute, highpressure cutoff , lowpressure }, speed 150 R.P.M. What will t
horsepower and steam consumption hypothetical and probable?
NOTE: S for 125 Ibs. =.315, for 5 Ibs. =.014.
Prob. 31. By graphical means find the (m.e.p.), of a 15X22x30in. crof:
compound engine, with 5 per cent clearance in each cylinder, if the receiver volume
8 cu.ft., initial pressure 125 Ibs. per square inch absolute, back pressure 10 Ibs. p
square inch absolute, highpressure cutoff f , lowpressure &, highpressure comprt
sion 40 per cent, lowpressure 20 per cent, highpressure crank following 90, logarithm
expansion.
Prob. 32. Show by a series of curves, assuming necessary data, the effect <
(m.e.p.) of cutoff, back pressure, clearance, and compression.
Prob. 33. Show by curves, how the indicated, and actual water rate, of an 18 X2:
in. engine with 5 per cent clearance, and running at 125 R.P.M. on an initial presst
of 125 Ibs. per square inch gage, and a back pressure of 10 Ibs. per square inch absolu
may be expected to vary with cutoff from ^o to f .
TABLES
395
TABLE XIII
PISTON POSITIONS FOR ANY CRANK ANGLE
FROM BEGINNING OF STROKE AWAY FROM CRANK SHAFT TO FIND PISTON POSITION FROM
DEADCENTER MULTIPLY STROKE BY TABULAR QUANTITY
Crank
Angle.
U
r
U 5
r
5
r
f 5 .5
f 6
i*
r
fs
'=9
T
5
.0014
.0015
.0015
.0016
.0016
.0016
.0017
.0019
10
.0057
.0059
.0061
.0062
.0063
.0065
.0067
.0076
15
.0128
.0133
.0137
.0140
.0142
.0146
.0149
.0170
20 ! .0228
.0237
.0243
.0248
.0253
.0260
.0265
.0302
25
.0357
.0368
.0379
.0388
.0394
.0405
.0413
.0468
30
.0513
.0531
.0545
.0556
.0565
.0581
.0592
.0670
35
.0698
.0721
.0740
.0754
.0767
.0787
.0801
.0904
40
.0910
.0939
.0962
.0981
.0997
.1022
.1041
.1170
45
.1152
.1187
.1215
.1237
.1256
.1286
.1308
.1468
50
.1416
.1458
.1491
.1518
.1541
.1576
.1607
.1786
55
.1713
.1759
. 1828"
.1827
.1853
.1892
.1922
.2132
60
.2026
.2079
.2122
.2157
.2186
.2231
.2295
.2500
65
.2374
.2431
.2477
.2514
.2545
.2594
.2630
.2886
70
.2730
.2794
.2844
.2885
.2929
.2973
.3013
.3290
75
.3123
.3187
.3239
.3282
.3317
.3372
.3414
.3705
80
.3516
.3586
.3642
.3687
.3725
.3784
.3828
.4132
85
.3944
.4013
.4068
.4113
.4151
.4210
.4254
.4564
90
.4365
.4437
.4495
.4547
.4580
:4641
.4686
.5000
95
.4816
.4885
.4940
, .4985
.5022
.5081
.5126
.5436
100
.5253
.5323
.5378
.5424
.5461
.5520
.5564
.5868
105
.5711
.5775
.5828
.5870
.5905
.5961
.6002
.6294
110
.6150
.6214
.6265
.6306
.6340
.6393
.6530
.6710
115
.6600
.6657
.6703
.6740
.6771
.6820
.6856
.7113
120
.7026
.7080
.7122
.7157
.7186
.7231
.7265
.7500
125
.7449
.7495
.7533
.7563
.7588
.7628
.7658
.7868
130
.7844
.7885
.7920
.7947
.7969
.8004
.8030
.8214
135
.8223
.8258
.8286
.8308
.8327
.8357
.8379
.8535
140
.8570
.8600
.8623
.8642
.8658
.8682
.8703
.8830
145
.8889
.8913
.8931
.8946
.8958
.8978
.8993
.9096
150
.9173
.9191
.9204
.9216
.9226
.9241
.9252
.9330
155
.9420
.9432
.9452
.9451
.9457
.9468
.9476
.9531
160
.9625
.9633
.9640
.9645
.9650
.9656
.9661
.9698
165
.9787
.9792
.9796
.9799
.9802
.9805
.9809
.9829
170
.9905
.9908
.9909
.9911
.9912
.9913
.9915
.9924
175
.9976
.9977
.9977
.9977
.9978
.9978
.9979
.9981
180
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
396
ENGINEERING THERMODYNAMICS
TABLE XIV
VALUES OF x FOR USE IN HECK'S FORMULA FOR MISSING WATER
Absolute
Steam Pressure.
X
Absolute
Steam Pressure.
X
Absolute
Steam Pressure.
X
170
70
297.5
165
393
1
175
75
304
170
397
2
179
80
310
180
405
3
183
85
316
185
409
4
186
90
321.5
190
413
6
191
95
327
K5
416.5
8
196
100
332.5
200
420
10
200
105
338
210
427
15
210
110
343
220
431
20
220
115
348
230
441
25
229
120
353
240
447.5
30
238
125
358
250
454
35
246
130
362.5
260
460.5
40
254
135
367
270
467
45
262
140
371.5
280
473
50
269.5
145
376
290
479
55
277
150
380.5
300
485
60
284
155
385
65
291
160
389
TABLE XV
SOME ACTUAL ENGINE DIMENSIONS
SIMPLE
7X9
7X15
16 X18
15X24
24 X3G
8X9
8X15
16X18
16 X24
26 X36
9X9
12 X15
17 X18
18 X24
26X38
5X10
13 X15
17X18
20 X24
28 X36
6^X10
14 X15
18 X18
22 X24
14 X42
8 X10 14X15
19 X18
24 X24
15 X42
9 X10
15 X15
20 X18
16X27
16 X42
10 X10
16 X15
29 X19
17^X27
18 X42
11 X10
17^X15
12 X20
10 X30
20 X42
9X10
11 X16
14 X20
12 X30
22 X>2
10X10
12 X16
18 X20
16 X30
24 X42
7JX12
13 X16
19 X20
18 X30
26 X42
8 X12
14iXl6
28 X20
18fX30
28 X^
8fX12
15 X16
21 X20
20 X30 . .
18 X48
9 X12
15iX16
22 X20
24 X30
20 X48
10 X12
16 X16
12 X21
22 X33
22 X48
11 X12
17 X16
13 X21
24 X33
24 X48
1UX12
18 X16
18X21
10 X36.
26 X48
12 X12
18JX17
20 X21
12 X36
28 X48
12X12
23 X17
20 X22
14 X36
24 X54
13 X12
26 X17
18 X24
16 X36
26 X54
14 X12
10 X18
10 X24
18 X36
28 X54
10 X14
11 X18
12 X24
20 X36
28 X60
11 X14
15 X18
14X24
22 X36
TABLES
397
TABLE XV. Continued
COMPOUND
NOTES: 1 to run condensing or noncondensing on initial pressure of 100150.
2 to run condensing or noncondensing on initial pressure of 100.
3 to run condensing or noncondensing on initial pressure of 125.
4 to run condensing or noncondensing on initial pressure of 90100.
5 to run condensing or noncondensing on initial pressure of 110130.
6 to run condensing or noncondensing on initial pressure of 140160.
7 to run condensing or noncondensing on initial pressure of 125
4J 8 X 6
13 18 X14
1
11 19 X18
3
14i26 X22
4
22 38 X33
3
6 10 X 6
13 20 X14
1,2
12 21 X18
3
18 32 X22
5
24 42 X33
3
7 13 X 8
7f13 X 15
3
13 22^X18
3
10 17^X24
3
18132^X36
3
6 12 X10
1
9 151X15
3
14 24 X18
1
11 19 X24
3
20 36 X36
3
7 12 X10
1
11 19 X15
4
15126^X18
3
12 18 X24
7
26^46 X36
3
8 12 X10
1
13 19 X15
5
16 24 X18
1
13 20 X24
7
28^50 X36
3
8f15X10
713X16
3
16 26 X18
14 22 X24
7
1425 X42
3
7 14 X10
1
9 15X16
3
16^28^X18
3
16^28X24
3
15126^X42
3
8 14 X10
1
10 17IX16
3
8 12 X20
7
1730X24
3
18^32^X42
3
915 Xll
2
11 19 X16
3
9 14 X20
7
22 38 X24
3
20 36 X42
3
713X12
3
11 22 X16
1
16 28 X20
1
24 42 X24
3
1628X48
3
9 151X12
3
12 21 X16
3
17 30 X20
1
12 21 X27
3
17^30^X48
3
19 14 X12
1
13 22 X16
1
18 28 X20
1
13 22X27
3
22 38 X48
3
10 16 X12
1
13 221X16
3
19 30 X20
1
16^28^X27
3
; 24 42 X48
3
10 18 X12
1
14 22 X16
2
19 30 X22
1
17130^X27
3
! 18132^X54
3
11 16 X12
1
14J25 X16
3
9 15X21
3
14125 X30
3
20 36 X54
3
9 18 X14
1
15 22 X16

12 21 X21
3
15126X30
3
26 46 X54
3
10 18 X14
1
1526X16
3
13 22X21
3
18 1 32X30
3
28^50 X54
3
10 17^X14
3
16 25 X16
2
14i26X21
3
20 36 X30
3
22 38 X60
3
11 19 X14
3
13 23 X17
4,6
15128^X21
3
28^50 X30
3
24 42 X60
3
11 18 X14
1
15 26 X17
4,6
18f32! X 21
3
30 54 X30
3
30 54 X60
3
2 18 X14
1
1629 X17
4
20 36 X21
3
16128^X33
3
3257 X60
3
12 20 X14
1
9 15X18
3
13 23 X22
5
17^X30^X33
3
34 60 X60
3
TRIPLE
NOTE: All condensing and to run of initial pressure as given.
SIZE.
P
SIZE.
P
SIZE.
P
10 15126X15
1118 30X20
200
250
27 43 { SJX39
\ oi
180
305082X48
254168X48
180
190
12 20 34X24
180
18 28M8X40
180
274575X54
190
12 19 32X24
190
22 37 63X42
180
284572X54
185
175
22 38 64X42
185
284675X54
180
12222 36X24
14 23 28X26
180
190
3253 {^X48
265
294783X54
325292X50
160
200
; 18 29 47X30
1624 41X30
200
180
35571^X48
265
3456100X60
3558 (~oX 60
200
190
18 30 50X30
16 25143X30
200
190
36 57  {^X48
295
\ 69
3457104X63
200
1624 41X30
180
28 45 72X48
180
CHAPTER IV
HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS
BETWEEN THE HEAT CONTENT OF SUBSTANCES AND THEIR PHYSICAL
CHEMICAL STATE,
1. Substances and Heat Effects Important in Engineering. It has been
shown in preceding chapters concerned with work in general and with the deter
mination of quantity of work that may be done in the cylinders by, or on
expansive fluids that
(a) Fluids originally at low may be put in a highpressure condition by
the expenditure of work;
(b) Fluids under high pressure may do work in losing that pressure.
That work may be done, fluids under pressure are necessary and that the
greatest amount of work may be done per unit of fluid the fluid itself must be
expansive, that is, it must be a gas or a vapor. Gases or vapors under pressure
are, therefore, prerequisites to the economical use of fluids for the doing of work,
and that this work may be done at the expense of heat or derived from heat, it
is only necessary that the heat be used to create the necessary primary con
dition of high pressure in vapors and gases. There are two general ways of
accomplishing this purpose first, to apply the heat to a boiler supplied with
liquid and discharging its vapor at any pressure as high as desired or as high
as may be convenient to manage; second, to apply the heat to a gas confined
in a chamber, raising its pressure if the chamber be kept at a fixed volume,
which is an intermittent process, or increasing the fluid volume if the size of the
chamber be allowed to increase, the fluid pressure being kept constant or not,
and this latter process may be intermittently or continuously carried out.
These two processes are fundamental to the steam and gas engines that are
the characteristic prime movers or power generators of engineering practice,
utilizing heat energy, and with the exception of waterwheels the sole commer
cially useful sources of power of the industrial world. Thus, the heating of
gases and the evaporation of liquids are two most important thermal processes
to be examined together with their inverse, cooling and condensation, and
necessarily associated in practical apparatus with the heating and cooling of
solid containers or associated liquids. From the power standpoint, the effects
of heat on solids, liquids, gases and vapors, both without change of state and with
change of state are fundamental, and the substances to be studied as heat carriers
do not include the whole known chemical world, but only those that are cheap
enough to be used in engineering practice or otherwise essential thereto. These
substances of supreme importance are, of course, air and water, with all their
398
HEAT AND MATTER 399
jiysical and chemical variations, next the fuels, coal, wood, oil, alcohol and
jmbustible gases, together with the chemical elements entering into them
 id the chemical compounds which mixed together may constitute them.
] Probably next in importance from the standpoint of engineering practice
je the substances and thermal processes entering into mechanical refriger
jion and ice making. There are but three substances of commercial importance
jsre ammonia, pure and in dilute aqueous solution, carbonic acid and air.
iae process of heating or cooling solids, liquids, gases and vapors, together
Hth solidification of water into ice, evaporation and condensation, fundamental
f power problems, are also of equal importance here, but there is added an
ilditional process of absorption of ammonia vapor in water and its discharge
om the aqueous solution.
Many are the practical applications of heat transfer or transmission, some
which call into play other "substances than those named. In the heating
 buildings there is first combustion with transfer of heat to wa,ter in boilers,
i)W of the hot water or steam produced to radiators and then a transfer of heat
^ the air of the room; in feedwater heaters, heat of exhaust steam warms
ater on its way to. the boilers; in economizers, heat of hot flue gases is trans
rred to boiler feed water; in steam superheaters, heat of hot flue gases is trans
rred to steam previously made, to raise its temperature, steam pipes, boiler
lirfaces and engine cylinders transfer heat of steam to the air which is opposed
r covering and lagging, in steam engine condensers heat of exhaust steam is
;ansf erred to circulation water; in cooling cold storage rooms and making ice,
i solution of calcium or sodium chloride in water is circulated through pipes
jid tanks and is itself kept cool in brine coolers in which the brine transfers
he heat absorbed in the rooms and tanks, to the primary substance ammonia
F carbonic acid and evaporates it.
While evaporation and condensation as processes are fundamental to the
tachinery and apparatus of both power and refrigeration, they also are of
aportance in certain other industrial fields. In the concentration of
>lutions to promote crystallization such, for example, as sugar, evapora
m of the solution and condensation of the distillate are primary processes
[fclso is the case in making gasolene and kerosene from crude oil, in the making
ftalcohol from a mash, and many other cases found principally in chemical
jianufacture. These are examples of evaporation and condensation in which
jttle or no gases are present with the vapor but there are other cases in which
! gas is present in large proportion, the thermal characteristics of which are
lifferent as will be seen later. Among these processes are: the humidification
; t moistening of air with water in houses and factories to prevent excessive
lin evaporation of persons breathing the air, excessive shrinkage of woodwork
to facilitate the manufacturing processes like tobacco working and thread
finning. Conversely, air may be too moist for the purpose, in which case it
fcried by cooling it and precipitating it$ moisture as rain or freezing it out as
m, and this is practiced in the Gayley process of operating blast furnaces, where
Mbess of moisture will on dissociating absorb heat of coke combustion and reduce
400 ENGINEERING THERMODYNAMICS
the iron output per ton of coke, and in the factories where, for example, collodior
is worked, as in tb~ manufacture of photographic films, with which moisture
seriously interferes. Of course, humidification of air by water is accomplishec
only by evaporation of water, and evaporation of water is only to be accomplishec
, by the absorption of heat, so that humidification of air by blowing it over watei
or spraying water into it must of necessity cool the water, and this is the prin
ciple of the cooling tower or cooling pond for keeping down the temperature
of condenser circulating water, and likewise the principle of the evaporative
condenser, in which water cooler and steam condenser are combined in one
The same process then, may serve to cool water if that is what is wanted, or tt
moisten air, when dry air is harmful, and may also serve to remove moistun
from solids like sand, crystals, fabrics, vegetable or animal matter to be reducee
to a dryer or a pulverized state.
There are some important examples of humidification in which the substance;
are not air and water, and one of these is the humidification of air by gasolene
or alcohol vapor to secure explosive mixtures for operating gas engines. Her<
the air vaporizes enough of the fuel, humidifying or carburetting itself to serv<
the purpose, sometimes without heat being specifically added and sometime,
with assistance from the hot exhaust. A somewhat similar action takes plac<
in the manufacture of carburetted water gas when the water gas having n<
illuminating value is led to a hot brick checkerwork chamber supplied with i
hydrocarbon oil, the vapor of which humidifies the gas, the heat of vaporizatioi
being supplied by the hot walls and regularly renewed as the process is inter
mittent. Of course, in this case some of the vapors may really decompos
into fixed gases, a peculiar property of the hydrocarbon fuels, both liquid an*
gaseous, and frequently leaving residues of tar, or soot, or both.
Finally, among the important processes there is to be noted that of gasifica
tion of solid and liquid fuels in gas producers and vaporizers, a process alsj
carried on in blast furnaces in which it is only an accidental accompanimen
and not the primary process. Some of the actions taking place in gas producei
are also common to the manufacture of coal gas, and coke, in retorts, beehiv
and byproduct ovens.
From what has been said it should be apparent that engineers are conceme
not with any speculations concerning the nature of heat but only with the kin
and quantity of effect that heat addition to, or abstraction from, substance
may be able to produce and not for all substances either. While this interee
is more or less closely related to philosophic inquiry, having for its object tb
development of all embracing generalizations or laws of nature, and to tl
relation of heat to the chemical and physical constitution of matter, subje<
matter of physical chemistry, the differences are marked, and a clearly define
field of application of laws to the solution of numerical problems dealing wit
identical processes constitutes the field of engineering thermodynamics.
It is not possible or desirable to take up and separately treat every sing
engineering problem that may rise, but on the contrary to employ the scientil
methods of grouping thermal processes or substance effects into types.
HEAT AND MATTER 401
Prob. 1. Water is forced by a pump through a feedwater heater and economizer
> a boiler where it is changed to steam, which in turn passes through a superheater
) a cylinder from which it is exhausted to a condenser. Which pieces of apparatus
lave to do with heat effects and which with work? Point out similarities and
jifferences of process.
Prob. 2. Air is passed over gasolene in a carburetter; the mixture is compressed,
jurned and allowed to expand in a gas engine cylinder. Which of the above steps have
h do with heat effects and which with work effects?
Prob. 3. In certain types of ice machines liquid ammonia is allowed to evaporate,
ic vapor which is formed being compressed and condensed again to liquid. Which of
ifese steps is a work phase and which a heat phase? Compare with Problem 1.
> Prob. 4. When a gun is fired what is the heat phase and what is the work
hase? Are they separate or coincident?
Prob. 6. Air is compressed in one cylinder, then it is cooled and compressed to
igher pressure and forced into a tank. The air in the tank cools down by giving
p heat to the atmosphere. From the tank it passes through a pipe line to a heater
ad then to an engine from which it is exhausted to the atmosphere. Which steps
: the cycle may be regarded as heat and which as work phases? Compare with
'roblem 2.
2. Classification of Heating Processes. Heat Addition and Abstraction
rith, or without Temperature Change. Qualitative Relations. That heat
pass from a hot to a less hot body if it gets a chance is axiomatic, so that a
>ody acquiring heat may be within range of a hotter one, the connection between
inem being, either immediate, that is they touch each other, or another body
nay connect them acting as a heat carrier, or they may be remote with no more
lovable connection than the hypothetic ether as is the case with the sun and
rth. A body may gain heat in other ways than by transfer from a hotter
>ody, for example, the passage of electrical current through a conductor will
eat it, the rubbing of two solids together will heat both or perhaps melt one,
lie churning of a liquid will heat it, the mixing of water and sulphuric acid will
reduce a hotter liquid than either of the components before mixture, the absorp
ion by water of ammonia gas will heat the liquid. All these and many other
imilar examples that might be cited have been proved by careful investigation
>artly experimental, and partly by calculation based on various hypotheses to
>e examples of transformation of energy, mechanical, electrical, chemical, into
le heat form. While, therefore, bodies may acquire heat in a great many
different concrete ways they all fall under two useful divisions:
(a) By transfer from a hotter body;
(6) By transformation into heat of some other energy manifestation.
One body may be said to be hotter than another when it feels so to the
iense of touch, provided neither is too hot or too cold for injury to the tissues,
Ir more generally, when by contact one takes heat from the other. Thus.
Seas of heat can scarcely be divorced from conceptions of temperature and the
definition of one will involve the other. As a matter of fact temperature as
indicated by any instrument is merely an arbitrary number located by some
lody on a scale, which is attached to a substance on which heat has some visible
402 ENGINEEEING THEEMODYNAMICS
effect. Temperature is then a purely arbitrary, though generally acceptec
number indicating some heat content condition on a scale, two points of whicii
have been fixed at some other conditions of heat content, and the scale spac
between, divided as convenient. Examination of heat effects qualitative!
will show how thermometers might be made or heat measured in terms of an
handy effect, and will also indicate what is likely to happen to any substanc
when it receives or loses heat. Some of the more common heat effects of variou
degrees of importance in engineering work are given below:
Expansion of Free Solids. Addition of heat to free solids will cause them t l
expand, increasing lengths and volumes. Railroad rails and bridges are longe
in summer than winter and the sunny side of a building becomes a little highe
than the shady side. Steam pipes are longer and boilers bigger hot, than cole
and the inner shell of brick chimneys must be free from the outer to permi
it to grow when hot without cracking the outer or main supporting stack bodj
Shafts running hot through lack of lubrication or overloading in comparative!
cool bearing boxes may be gripped tight enough to twist off the shaft or merel;
score the bearing.
Stressing of Restrained Solids. A solid being heated may be restraine
in its tendency to expand, in which case there will be set up stresses in the mate
rial which may cause rupture. Just as with mechanically applied loads, bodie
deform in proportion to stress up to elastic limit, as stated by Hooke's law, s
if when being heated the tendency to expand be restrained the amount .c
deformation that has been prevented determines the stress. A steam pip
rigidly fixed at two points when cold will act as a long column in compressio
and buckle when hot, the buckling probably causing a leak or rupture. If fixe
hot, it will tend to shorten on cooling and being restrained will break something
Cylinders of gas engines and air compressors are generally jacketed with wate
and becoming hot inside, remaining cold outside, the inner skin of the mete
tends to expand while the outer skin does not. One part is, therefore, in tensio
and the other in compression, often causing cracks when care in designing i
not taken and sometimes in spite of care in large gas engines.
Expansion of Free Liquids. Heating of liquids will cause them to expan
just as do solids, increasing their volume. Thus, alcohol or mercury in gla^
tubes will expand and as these liquids expand more than the glass, a tube whic
was originally full will overflow when hot, or a tube of very small bore attache
to a bulb of cold liquid will on heating receive some liquid; the movement c
liquid in the tube if proportional to the heat received will serve as a thermometei
If the solid containing the liquid, expanded to the same degree as the liqui
there would be no movement. Two parts of the same liquid mass may b
unequally hot and the hotter having expanded will weigh less per cubic foo
that is, be of less density. Because of freedom of movement in liquids the lightc
hot parts will rise and the cooler heavy parts fall, thus setting up a circulatior
the principle of which is used in hot water heating systems, the hot water froi
the furnace rising to the top of the house through one pipe and cooling on 11
downward path through radiators and return pipe. In general then, liquie
HEAT AND MATTER 403
decrease in density on heating and increase in density on cooling, but a most
important exception is water, which has a point of maximum density just
above the freezingpoint, and if cooled below this becomes not heavier but lighter.
Consequently, water to be cooled most rapidly should be cooled at first from
the top and after reaching this point of maximum density, from the bottom, if it
is to be frozen.
Rise of Pressure in Confined Liquids. When liquids are restrained from
expanding under heating they suffer a rise of pressure which may burst the
containing vessel. For this reason, hot water heating systems have at their
highest point, open tanks, called expansion tanks, which contain more water
when the system is hot than when cold, all pipes, radiators and furnaces being
constantly full of water. Should this tank be shut off when the water is cold
something would burst, or joints leak, before it became very hot.
Expansion of Free Gases. Just as solids and liquids when free expand under
heating, so also do gases and on this principle chimneys and house ventilation
systems are designed. The hot gases in a chimney weigh less per cubic foot
than cooler atmospheric air; they, therefore, float as does a ship on water,
the superior density of the water or cold gas causing it to flow under and
lift the ship or hot gas, respectively. Similarly, hotair house furnaces and
ventilating systems having vertical flues supplied with hot air can send it upward
by simply allowing cold air to flow in below and in turn being heated flow up
and be replaced.
Rise of Pressure in Confined Gases. Gases when restrained from expanding
under heat reception will increase in pressure just as do liquids, only over greater
ranges, and as does the internal stress increase in solids when heated under
restraint. It is just this principle which lies at the root of the operation of
guns and gas engines. Confined gases are rapidly heated by explosive combus
tion and the pressure is thus raised sufficiently to drive projectiles or pistons
in their cylinders.
Melting of Solids. It has been stated that solids on being heated expand
but it should be noted that this action cannot proceed indefinitely. Continued
heating at proper temperatures will cause any solid to melt or fuse, and the pre
viously rising temperature will become constant during this change of state.
IThus, melting or fusion is a process involving a change of state from solid
to liquid and takes place at constant temperature. The tanks or cans of ice
making plants containing ice and water in all proportions retain the same
temperature until all the water becomes ice, provided there is a stirring or cir
[ culation so that one part communicates freely with the rest and provided also
the water is pure and contains no salt in solution. Impure substances, such as
I liquid solutions, may suffer a change of temperature at fusion or solidification.
; For pure substances, melting and freezing, or fusion and solidification, are
 constant temperature heat effects, involving changes from solid to liquid, or
liquid to solid states.
Boiling of Liquids. Ebullition. Continued heating of solids causes fusion,
and similarly continued heating of liquids causes boiling, or change of state from
404 ENGINEERING THERMODYNAMICS
liquid to vapor, another constant temperature process just what temperature,
will depend on the pressure at the time. So constant and convenient is this
temperature pressure relation, that the altitude of high mountains can be found
from the temperature at which water boils. The abstraction of heat from a
vapor will not cool it, but on the contrary cause condensation. Steam boilers
and ammonia refrigerating coils and coolers are examples of evaporating appara
tus, and house heating radiators and steam and ammonia condensers of con
densing apparatus.
Evaporation of Liquids; Humidification of Gases. When dry winds blow
over water they take up moisture in the vapor form by evaporation at any
temperature. This sort of evaporation then must be distinguished from ebul
lition and is really a heat effect, for without heat being added, liquid cannot
change into vapor; some of the necessary heat may be supplied by the water
and some by the air. This process is general between gases and liquids and is
the active principle of cooling towers, carburetters, driers of solids like wood
kilns. The chilling of gases that carry vapors causes these to condense in part.
As a matter of fact it is not necessary for a gas to come into contact to produce
this sort of evaporation from a liquid, for if the liquid be placed in a vacuum
some will evaporate, and the pressure finally attained which depends on the tem
perature, is the vapor pressure or vapor tension of the substance, and the amount
that will so evaporate is measured by this pressure and by the rate of removal
of that which formed previously.
Evaporation of Solids. Sublimation. Evaporation, it has been shown, may
take place from a liquid at any temperature, but it may also take place directly
from the solid, as ice will evaporate directly to vapor either in the presence
of a gas or alone. Ice placed in a vacuum will evaporate until the vapor tension
is reached, and it is interesting to note that the pressure of vapors above theirij
solids is not necessarily the same as above their liquids at the same temperature,
though they merge at the freezingpoint. This is the case with icewater
water vapor.
Change of Viscosity. Heating of liquids may have another effect measured
by their tendency to flow, or their viscosity. Thus, a thick oil will flow easier
when heated, and so also will any liquid. If, therefore, the time for a giver 
quantity to flow through a standard orifice under a given head or pressure bt
measured, this time, which is the measure of viscosity, will be less for any liquk
hot, than cold, for the same liquid. Viscosity then decreases with heat additior
and temperature rise.
Dissociation of Gases. When gases not simple are heated and the heatinj
continued to very high temperatures, they will split up into their elements o
perhaps into other compound gases. This may be called decomposition or, better
dissociation, and is another heat effect. Thus, the hydrocarbon C2H 4 wil
split up with solid carbon soot C and the other hydrocarbon CH4 and steal)
H2O into hydrogen and oxygen. This is not a constant temperature process
but the per cent dissociated increases as the temperature rises.
Dissociation of Liquids. Similar to the dissociation of gases receiving heat a
HEAT AND MATTER 405
high temperature is the decomposition of some liquids in the liquid state, notably
the fuel and lubricating oils, or hydrocarbons which are compounds of H and C
in various proportions, each having different properties. Sometimes these
changes of H and C groupings from the old to the new compounds under the
influence of heating will be at constant and at other times at varying tempera
tures; sometimes the resulting substances remain liquid and sometimes soot
or C separates out, and this is one of the causes for the dark color of some
cylinder oils.
Absorption of Gases in Liquids. Liquids will absorb some gases quite freely;
thus, water will absorb very large quantities of ammonia, forming aqua ammonia.
Addition of heat will drive off this gas so that another heat effect is the expul
sion of gases in solution. Use is made of this industrially in the absorption
system of ammonia refrigeration.
Solubility of Solids in Liquids. The heating of liquids will also affect their
solubility for solid salts; thus, a saturated solution of brine will deposit crystals
on heat abstraction and take them back into solution on heat addition.
Certain scaleforming compounds are thrown down on heating the water in
tended for boilers, a fact that is made use of in feedwater heating purifiers;
for these salts increase of temperature reduces solubility. In general then heat
addition affects the solubility of liquids for solid salts.
Chemical Reaction. Combustion. If oxygen and hydrogen, or oxygen and
carbon, be heated in contact, they will in time attain an ignition temperature at
which a chemical reaction will take place with heat liberation called combus
tion, and which is an exothermic or heatfreeing reaction. Another and
different sort of reaction will take place if C02 and carbon be heated together,
for these will together form a combustible gas, CO, under a continuation of heat
reception. This is an endothermic or heatabsorbing reaction. Neither of
these will take place until by heat addition the reaction temperature, called
ignition temperature for combustion, has been reached.
Electrical and Magnetic Effects. Two metals joined together at two separate
points, one of which is kept cool and the other heated, will be found to carry
an electric current or constitute a thermoelectric couple. Any conductor
carrying an electric current will on changing temperature suffer a change of
resistance so that with constant voltage more or less current will flow; this is
a second electrical heat effect and like the former is useful only in instru
ments indicating temperature condition. A fixed magnet will suffer a change
of magnetism on heating so that heat may cause magnetic as well as electric
effects.
These heat effects on substances as well as some others of not so great engi
neering importance may be classified or grouped for further study in a variety
of ways, each serving some more or less useful purpose.
Reversible and Nonreversible Processes. There may be reversible and
nonreversible thermal processes, when the process may or may not be con
sidered constantly in a state of equilibrium. For example, as heat is applied
to boiling water there is a continuous generation of vapor in proportion to the
406 ENGINEERING THERMODYNAMICS
heat received; if at any instant the heat application be stopped the evapo
ration will cease and if the flow of heat be reversed by abstraction, condensa
tion will take place, indicating a state of thermal equilibrium in which the
effect of the process follows constantly the direction of heat flow and is con
stantly proportional to the amount of heat numerically, and in sign, of direction.
As an example of nonreversible processes none is better than combustion, in
which the chemical substances receive heat with proportional temperature rise
until chemical reaction sets in, at which time the reception of heat has no fur
ther relation to the temperatures, because of the liberation of heat by com
bustion which proceeds of itself and which cannot be reversed by heat
abstraction. Even though a vigorous heat abstraction at a rate greater than
it is freed by combustion may stop combustion or put the fire out, no amount
of heat abstraction or cooling will cause the combined substances to change
back into the original ones as they existed before combustion. The effect of
heat in such cases as this is, therefore, nonreversible.
Constant and Variable Volume or Density. When gases, liquids or solids
are heated they expand except when prevented forcibly from so doing, and as
a consequence they suffer a reduction of density with the increase of volume;
this is, of course, also true of changing liquids to their vapors. It should be
noted that all such changes of volume against any resistance whatever, occur
with corresponding performance of some work, so that some thermal processes
may directly result in the doing of work. Heating accompanied by no volume
change and during which restraints are applied to keep the volume invariable,
cannot do any work or suffer any change of density, but always results in change
of pressure in liquids, gases and vapors and in a corresponding change of internal
stress in solids.
Constant and Variable Temperature Processes. Another useful division, and
that most valuable in the calculation of relations between heat effect and heat
quantity, recognizes that some of the heating processes and, of course, cooling,
occur at constant temperature and others with changing temperature. For
example, the changes of state from liquid to solid, and solid to liquid, or freezing
and fusion, are constant temperature processes in which, no matter how much
heat is supplied or abstracted, the temperature of the substance changing state is
not affected, and the same is true of ebullition and condensation, or the changing
of state from liquid to vapor, and vapor to liquid. These latter constant
temperature processes must not be confused with evaporation, which may
proceed from either the solid or liquid state at any temperature whether constant
or not.
Prob. 1. From the time a fire is lighted under a cold boiler to the time steam
first comes off, what heat effects take place?
Prob. 2. What heat effects take place when a piece of ice, the temperature of
which is 20 F., is thrown onto a piece of redhot iron?
Prob. 3. What heat effects must occur before a drop of water may be evaporated
from the ocean, and fed back into it as snow?
Prob. 4. What heat changes take place when soot is formed from coal or oil?
HEAT AND MATTER 407
Prob. 5. In a gas producer, coal is burned to C0 2 , which is then reduced to CO.
team is also fed to the producer, and H and O formed from it. Give all the heat
lects which occur.
Prob. 6. By means of what heat effects have you measured temperature changes,
r have known them to be measured?
Prob. 7. When the temperature changes from 40 F. to 20 F., give a list of all
eat effects you know that commonly occur for several common substances. Do the
une for a change in the reverse direction.
Prob. 8. If a closed cylinder be filled with water it will burst if the temperature
je lowered or raised sufficiently. What thermal steps occur in each case?
; Prob. 9. If salt water be lowered sufficiently in temperature, a cake of fresh ice
jad a rich salt solution will be formed. State the steps or heat effects which occur
tiring the process.
'homson " every kind
408 ENGINEERING THERMODYMAMICS
of a double metallic bar, often brass and iron, consisting of a piece of e*
fastened to the other to form a continuous strip. The two metals are expam
by the temperature different amounts causing the strip to bend under heati
There are also in use electric forms for all temperatures, and these are
only reliable ones for high temperatures, both of the couple and resistai
types except one dependent on the color of a high temperature body, blj
when cold. That most useful and common class involving the inter
pendence of pressure and temperature, or volume and temperature, of a fl
is generally found in the form of a glass bulb or its equivalent, to wh
is attached a long, narrow glass tube or stem which may be open or closed
the end; open when the changes of fluid volume at constant pressure are
be observed and closed when changes of contained fluid pressure at const;
restrained volume are to be measured as the effect of temperature chan
For the fluid there is used most commonly a liquid alone such as mercury,
a gas alone such as air; though a gas may be introduced above mercury i
there may be used a liquid with its vapor above. When the fluid is a liqi
such as mercury, in the common thermometer, the stem is closed at the end
that the mercury is enclosed in a constantvolume container or as nearly so
the expansion or deformation of the glass will permit, which is not filled w
mercury, but in which a space in the stem is left at a vacuum or filled wit]
gas under pressure, such as nitrogen, to resist evaporation of the mercury
high temperatures. Gasfilled mercury thermometers, as the last form is call
are so designed that for the whole range of mercury expansion the press
of the gas opposing it does not rise enough to offer material resistance to
expansion of mercury or to unduly stress the glass container. It should
noted that mercury thermometers do not measure the expansion of mere!
alone, but the difference between the volume of mercury and the glass 'en vekj
but this is of no consequence so long as this difference is in proportion to
expansion of the mercury itself, which it is substantially, with proper gi
composition, when the range is not too great. Such thermometers indid
temperature changes by the rise and fall of mercury in the stem, and any nun: :
cal value that may be convenient can be given to any position of the merer
or any change of position. Common acceptance of certain locations of the s I'j
number, however, must be recognized as rendering other possible ones unnej*
sary and so undesirable. Two such scales are recognized, one in use with m
units, the centigrade, and the other with measurements in English units,
Fahrenheit, both of which must be known and familiar, because of the freq
necessity of transformation of numerical values and heat data from one syi=
to the other. To permit of the making of a scale, at least two points/mus b
fixed with a definite number of divisions between them, each called one de t
The two fixed points are first, the position of the mercury when the thermom
is in the vapor of boiling pure water at sea level, or under the standard ati
pheric pressure of 29.92" = 760 mm. of mercury absolute pressure,
second, the position of the mercury when the thermometer is surrounde
melting ice at the same pressure. These are equivalent to the boiling or
HEAT AND MATTER
409
densation, and melting or freezingpoints, of pure water at one atmosphere
pressure. The two accepted thermometer scales have the following character
istics with respect to these fixed points and division between them :
THERMOMETER SCALES
Pure Water
Freezingpoint,
at one atm. pr.
Pure Water
Boilingpoint,
at one atm. pr.
Number of Equal Divisions
Between Freezing and Boiling.
Centigrade scale
Fahrenheit scale
32
100
212
100
180
From this it appears that a degree of temperature change is on the centigrade
scale, jJQ of the linear distance between the position of the mercury surface
at the freezing and boilingpoints of water, and on the Fahrenheit scale. T J^ of
the same distance. From this the relation between a degree temperature change
for the two scales can be given.
One degree temperature 1 _180_9_
change centigrade j 100 5
of one degree temperature
change Fahrenheit;
or
One degree temperature ) _5 f of one degree temperature
change Fahrenheit J 9 { change centigrade.
It is also possible to set down the relation between scale readings, for when the
temperature is C., it is 32 F., and when it is 100 C. it is (180+32) =212 F.,
so that
g
Temperature Fahrenheit = 32 +^ (Temperature centigrade),
or
. Temperature centigrade = ^ (Temperature Fahrenheit 32).
For convenience of numerical work tables are commonly used to transform
temperatures from one scale to the other and such a transformation is shown
in a curve, Fig. 119, and in Table XXIX at end of the Chapter.
By reason of the lack of absolute proportionality between temperature
and effect, other fixed points are necessary, especially at high temperatures,
and the following of Table XVI have been adopted by the U. S. Bureau of Stand
ards and are considered correct to within 5 C., at 1200 C.
410
ENGINEERING THERMODYNAMICS
100 i
JO
90 
85  ?
75 /
70 /
20
5 i:
o
66
60 .
46   I
i il
f/.
!=;;[[=!===l===!===!in=:i5!= J;=====!N iitSliiillnHii
L M Degrees
S g 2 8 S S 1 I S Fahrenheit
to . . *
i i i i i i i i i j 320 1 I 1 1 1 1 1 1 1 '
_2 ._
: ::z :: sisljJ
^  310 
\\_\\ ~ 305
 300     ^ 
 295  ::::;;
 290    T  
285 H
u 100 140 i
::?: 2*0 ::
2 I
/  275 
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 lik I4J 1 1 1 1 1 1 1 1 1 1 1 1 1 fils Htl 1 1 1 1 1 1 1 1 1 1 1 Ilils 1 1 1 1 1 1 1 1 1 LILiy^
1 1 1 1 L 1 1 1 1 1 1 1 I s
^aquaaq^ saaaSaa s;  "T:"3I qi S ^s
._ _.  _  
tf 860
10 760
LO 000
:ii:ni;2
: /
86 
16 
. i
10  /
20
^
/
70 y^
70
10 20
i i i I i i i i i MI 160  
1 . . . f
''' 1~ 155   '.'.'.]'.
 "? 160 
\\r\\\ / 
( 145   j  
60 7
lit
130
130  
m l
:::z!::: 12( :: :::?:
^ 11., Z
y no ]/
105^ 
30 40
 270  
 265  7
 260  
 265 :;;j;;;
 250^
J 80 120 U
? 246
 Z
/ 240  
::::::::: 235:::::::xi
/  230
_. . _.. 2 
 225      Z  
. 220 ?
 216  2"
U 60 100 1
Degrees Centigrade
Z  850
 800 
 750  '.~~fl
 700 _/_
3iO I
40
i i i i i i [ i i 1 1 650 
 _ _ . / _
__Z 600
/  550 /
i/ IN \y\
'f\<  w J^t
t: o 
 420 
370 /
10 120 100 2
LO 560
10 460
300
LO 260
FIG. 119. Granhical Relation between Centierrade and Fahrenheit Thermometer Scales.
HEAT AND MATTER
411
TABLE XVI
FIXED TEMPERATURES
U. S. BUREAU OF STANDARDS
Temperature,
Temperature,
Determined by the Point at which
232
449
Liquid tin solidifies
327
621
Liquid lead solidifies
419.4
787
Liquid zinc solidifies
444.7
832.5
Liquid sulphur boils
630.5
1167
Liquid antimony solidifies
658
1216
Liquid aluminum, 97.7% pure,
solidifies
1064
1947
Solid gold melts
1084
1983
Liquid copper solidifies
1435
2615
Solid nickel melts
1546
2815
Solid palladium melts
1753
3187
Solid platinum melts
Thermometers in which a liquid and its vapor exist together, depend on a
operty to be noted in detail later, the relation of vapor pressure to tempera
re and its independence of the volume of vapor. So long as any vapor exists
bove the liquid the temperature will depend only on the pressure of that vapor
o that such thermometers will indicate temperature by the pressure measure
aent, after experimental determination of this pressuretemperature relation
f vapors. Conversely, temperature measurements of vapors by mercury ther
nometers will lead to pressure values, and at the present time some steam
lants are introducing mercury thermometers on the boilers and pipe lines, in
lace of the proverbially inaccurate pressure gages.
Gas thermometer, is the name generally applied to the class in which the
uid is a gas, whether air, hydrogen, nitrogen or any other, and whether the
ressure is measured for a fixed contained volume, or the Volume measured
rhen acted on by a constant pressure. These gas thermometers are so bulky
s to be practically useless in ordinary engineering work and are only employed
a standards for comparison and for tests of extraordinary delicacy in investi
ation work. They give much larger indications than mercury thermometers
>ecause the changes of gas volume under constant pressure are far greater
ban for mercury or any other liquid. Regnault was the first to thoroughly
avestigate air thermometers and reported that the second form, that of constant
as volume with measurement of pressure, was most useful.
Using the centigrade scale, fixing freezing point at C., and making the
prresponding pressure po, atmospheric at this point, and reading at 100 C.
mother pressure pioo, he found experimentally a relation between these two
'pressures and the temperature corresponding to any other pressure p, as
ven by the empiric formula,
(601)
412 ENGINEERING THERMODYNAMICS
He also determined the pressure at the boilingpoint to be related to the pressu]
at the freezingpoint, by
pioo = 1.3665 po,
which on substitution gives
== 272.85 l ...... (60!
.3665j>o \po /
This constant, 272.85, the reciprocal of which is .003665, is, of course, the pressui
increase factor per degree C. rise of temperature for a gas held at const ai
volume, received extended investigation and it was found that it had abo.
the same value applied to the other type of thermometer in which gas volum<
are measured at constant pressure. This was true even when the pressu]
used was anything from 44 to 149 cm. of mercury, though it is reporte
that for 44 cm. pressure the value 272.98, and for 149 cm. pressure, the vain
272.7, seemed closer. For hydrogen it was found that the constant was sul
stantially the same as for air, while for carbonic acid it was 270.64, and whi
the hydrogen thermometers agreed with the air over the whole scale, showir
proportional effects, this was hardly true of carbonic acid. Such Uncertain!
in the behavior of these thermometers and in the fixing of the constants w;
traced to the glass in some cases, but there still remained differences charg
able only to the gases themselves. Comparison of the air with mercury the
mometers showed that there was not a proportional change with the temperatu
and that temperatures on the two, consistently departed.
Examination of Eq. (602) , giving the relation between two temperatures ar
the corresponding gas pressures, will show a most important relation. If in E
(602), the pressure be supposed to drop to zero and it is assumed th
the relations between pressure and temperature hold, then when p =
=272.85. This temperature has received the name of the absolute ze
and may be defined as the temperature at which pressure disappears or becom
zero at constant volume, and correspondingly, at which the volume also d
appears, since it was found that similar relations existed between volume ai
temperature at constant pressure. Calling temperature on a new scale begi
ning 272.85 below the centigrade zero by the name absolute temperature
then
[Absolute temperature 1 f Scale temperature]
centigrade centigrade J
.' '
As this constant or absolute temperature of the centigrade scale zero, is
experimental value, it is .quite natural to find other values presented by diff<
ent investigators, some of them using totally different methods. One of th(
methods is based on the temperature change of a gas losing pressure withe
doing work, generally described as the porous plug experiment, and the resu
HEAT AND MATTER 413
s the JouleThomson effect, and another is based on the coefficient of expansion
f gases being heated. Some of these results agreed exactly with Regnault's
alue for hydrogen between C. and 100 C. for which he gave 273 C.=
491.4 F. Still other investigations continued down to the last few years
ielded results that tend to change the value slightly to between 491.6 F.,
nd 491.7 F., and as yet there is no absolute agreement as to the exact value.
a engineering problems, however, it is seldom desirable or possible to work
o such degrees of accuracy as to make the uncertainty of the absolute zero a
.latter of material importance, and for practical purposes the following values
my be used with sufficient confidence for all but exceptional cases which are
i be recognized only by experience.
.
[Centigrade = 273)
Absolute Temperature (T) j Fahrenheit = 460 [ +^ ca ^ e Temperature (Q.
great accuracy is important it is not possible at present to get a better
''ahrenheit value than 459.65, the mean of the two known limits of 459.6 and
59.7, though Marks and Davis in their Steam Tables have adopted 459.64, which
3 very close to the value of 459.63 adopted by Buckingham in his excellent
bulletin of U. S. Bureau of Standards and corresponding to 273.13 on the
tigrade scale.
These experiments with the gas thermometers, leading to a determination
f temperature as a function of the pressure change of the gas held at constant
olume, or its volume change when held at constant pressure, really supply a
'finition of temperature which before meant no more than an arbitrary number,
.d furnished a most valuable addition to the generalization of relations between
at content of a body and its temperature or physical state.
A lack of proportionality between thermometer indication and temperature,
as already been pointed out, and it is by reason of this that two identical ther
omcters, or as nearly so as can be made, with absolute agreement between
,ter boiling and freezingpoints, will not agree at all points between, nor will
e best constructed and calibrated mercury thermometers agree with a similarly
1 gas thermometer.
The temperature scale now almost universally adopted as standard is that
I the constant volume hydrogen gas thermometer, on which the degree F.
lone onehundredandeightieth part of the change in pressure of a fixed
blume of hydrogen between melting pure ice, and steam above boiling pure
Jlter, the initial pressure of the gas at 32 being 100 cm. =39.37 ins. Hg. A
rcury in glass thermometer indication is, of course, a measure of the proper
of the mercury and glass used, and its F. degree of temperature is defined
parallel with the above as one onehundredandeightieth part of the volume
the stem between its indications at the same two fixed points. A comparison
the hydrogen thermometer and two different glasses incorporated in mercury
rmometers is given below, Table XVII, from the Bulletin of the U. S.. Bureau
Standards, v. 2, No. 3, by H. C. Dickinson, quoting Mahlke, but it must be
414
ENGINEERING THERMODYNAMICS
remembered that other glasses will give different results ana even differer
thermometers of the same glass when not similarly treated.
TABLE XVII
FAHRENHEIT TEMPERATURES BY HYDROGEN AND MERCURY
THERMOMETERS
Temperature by
Hydrogen
Thermometer.
Difference in
Reading by
Mercurv in Jena
59" Glass.
Difference in
Reading by
Mercury in 16"
Jena Glass.
Temperature by
Hydrogen
Thermometer.
Difference in
Reading by
Mercury in Jena
59" Glass.
Difference in
Reading by
Mercury in 16'
Jena Glass.
32
617
+ 10.6
212
662
+ 16.6
302
 .18
707
+18.7
392
+ 1.3
+ .072
752
+24.6
428
+ .39
797
+28.2
464
+ .83
842
+38.3
500
+1.79
887
+41.4
536
+2.4
932
+50.0
572
+3.53
Useful comparisons of air, hydrogen, nitrogen, and other gases, with alcoh<
and mercury, in various kinds of glass, are given in the LandoltBornsteii
Meyerhoffer, and in the Smithsonian Physical Tables, but are seldom neede
for engineering work.
One sort of correction that is often necessary in mercury thermometi
work is that for stem immersion. Thermometers are calibrated as a rule wi1
the whole stem immersed in the melting ice or the steam, but are ordinari J
used with part of the stem exposed and not touching the substance whose ter
perature is indicated. For this condition the following correction is recoi
mended by the same Bureau of Standards Bulletin :
Stem correction = .000088 n (t ti) F
When n = number of degrees exposed;
t = temperature indicated Fahrenheit degrees;
ti =mean temperature of emergent stem itself, which must necessarii
be estimated and most simply by another thermometer next i*
it, and entirely free from the bath.
Prob. 1. What will be the centigrade scale and absolute temperatures, for it
following Fahrenheit readings? 25, 25, 110, 140, 220, 263 scale, and 30(,
460, 540, 710, 2000 absolute.
Prob. 2. What will be the Fahrenheit scale and absolute temperatures, for the folk I
ing centigrade readings? 20, 10, 45, 80, 400, 610 scale, and 200, 410, 65 j,
810, 2500 absolute.
Prob. 3. By the addition of a certain amount of heat the temperature of L
quantity of water was raised 160 F. How many degrees C. was it raised?
HEAT AND MATTER 415
Prob. 4. To bring water from C. to its boilingpoint under a certain pressure
required a temperature rise of 150 C. What was the rise in Fahrenheit degrees?
Prob. 6. For each degree rise Fahrenheit, an iron bar will increase .00000648 of
its length. How much longer will a bar be at 150 C. than at C.? At 910 C.
absolute than at 250 C. absolute?
Prob. 6. The increase in pressure for S0 2 for a rise of 100 C. is given as .3845 at
constant volume. What would have been absolute zero found by Regnault had he
used S02 rather than air?
Prob. 7. A thermometer with a scale from 40 F. to 700 F. is placed in a thermome
ter well so that the 200 mark is just visible. The temperature as given by the
i thermometer is 450. If the surrounding temperature is 100 F., what is true tempera
ture in the well?
4. Calorimetry Based on Proportionality of Heat Effects to Heat Quantity.
 Units of Heat and Mechanical Equivalent. Though it is generally recognized
from philosophic investigations extending over many years, that heat is one
manifestation of energy capable of being transformed into other forms such
its mechanical work, electricity or molecular arrangement, and derivable from
them through transformations, measurements of quantities of heat can be made
without such knowledge, and were made even when heat was regarded as a
substance. It was early recognized that equivalence of heat effects proved
effects proportional to quantity; thus, the melting of one pound of ice can cool
a pound of hot water through a definite range of temperature, and can cool
wo pounds through half as many degrees, and so on. The condensation of
i pound of steam can warm a definite weight of water a definite number of
iegrees, or perform a certain number of pounddegrees heating effect in water.
3o that taking the pounddegree of water as a basis the ratio of the heat liberated
iy steam condensation to that absorbed by ice melting can be found. Other
ubstances such as iron or oil may suffer a certain number of pound degree
hanges and affect water by another number of pounddegrees. The unit
>f heat quantity might be taken as that which is liberated by the condensation
f a pound of steam, that absorbed by the freezing of a pound of water, that to
fcise a pound of iron any number of degrees or any other quantity of heat
ffect. The heat unit generally accepted is, in metric measure, the calorie,
r the amount to raise one kilogramme of pure water one degree centigrade,
in English units, the British thermal unit, that necessary to raise one pound
jf water one degree Fahrenheit. Thus, the calorie is the kilogramme degree
entigrade, and the British thermal unit the pound degree Fahrenheit, and the
Itter is used in engineering, usually abbreviated to B.T.U. There is also
occasionally used a sort of cross unit called the centigrade heat unit, which is
he pound degree centigrade.
The relation between these is given quantitatively by the conversion table
t the end of this Chapter, Table XXX.
All the heat measurements are, therefore, made in terms of equivalent
rater heating effects in pound degrees, but it must be understood that a water
ound degree is not quite constant. Careful observation will show that the
416 ENGINEERING THERMODYNAMICS
melting of a pound of ice will not cool the same weight of water from 200 F.
to 180 F., as it will from 60 F. to 40 F., which indicates that the heat capacity
of water or the B.T.U. per pounddegree is not constant. It is, therefore,
necessary to further limit the definition of the heat unit, by fixing on some
water temperature and temperature change, as the standard, in addition to the
selection of water as the substance, and the pound and degree as units of capacity.
Here there has not been as good an agreement as is desirable, some using
4 C. = 39.4 F. as the standard temperature and the range onehalf degree
both sides; this is the point of maximum water density. Others have used 15
C. = 59 F. as the temperature and the range onehalf degree both sides; still
others, one degree rise from freezing point C. or 32 F. There are good
reasons, however, for the most common presentday practice which will prob
ably become universal, for taking as the range and temperatures, freezing
point to boilingpoint, and dividing by the number of degrees. The heat unit
so defined is properly named the mean calorie or mean British thermal unit;
therefore,
Mean calorie = (amount of heat to raise 1 Kg. water from C. to 100 C.),
1UU
Mean B.T.U.rgg (amount of heat to raise 1 Ib. water from 32 F. to 212 F.)
In terms of the heat unit thus defined, the amount of heat per degree tern
perature change is variable over the scale, but only in work of the most accurat<
character is this difference observed in engineering calculations, but in accurat
work this difference must not be neglected and care must be exercised in usinj
other physical constants in heat units reported by different observers, to be sur
of the unit they used in reporting them. It is only by experience that judgmen.
can be cultivated in the selection of values of constants in heat units reporte
for various standards, or in ignoring differences in standards entirely. Th;
great bulk of engineering work involves uncertainties greater than these diffei (
ences and they may, therefore, be ignored generally.
By various experimental methods, all scientifically carried out and exetndin
over sixty years, a measured amount of work has been done and entirely coi
verted into heat, originally by friction of solids and of liquids, for the dete
mination of the footpounds of work equivalent to one B. T. U., when tl
conversion is complete, that is, when all the work energy has been converted in
heat. This thermophysical constant is the mechanical equivalent of he<
Later, indirect methods have been employed for its determination by calcu)
tion from other constants to which it is related. All of these experimer
have led to large number of values, so that it is not surprising to find doubt
to the correct value and different values are used even by recognized autho
ties. The experiments used include:
HEAT AND MATTER 417
v
(a) Compression and expansion of air; Joule.
(b) Steam engine experiments, comparing heat in supplied and exhausted
steam; Hirn.
(c) Expansion and contraction of metals; Edlund and Haga.
(d) Specific volume of vapor; Perot.
(e) Boring of metals; Rumford and Him.
(/) Friction of water; Joule and Rowland.
(g) Friction of mercury; Joule.
(h) Friction of metals; Hirn, Puluj, Sahulka.
(i) Crushing of metals; Hirn.
(j) Heating of magneto electric currents; Joule.
(k) Heating of disk between magnetic poles; Violle.
(I) Flow of liquids (water and mercury) under pressure; Hirn, Bartholi.
(m) Heat developed by wire of known absolute resistance; Quintus Icilius,
Weber, Lenz, Joule, Webster, Dieterici.
(ri) Diminishing the heat contained in a battery when the current produces
work; Joule, Favre.
(o) Heat developed in, and voltage of Daniell cells; Weber, Boscha, Favre
and Silberman, Joule.
(p) Combination of electrical heating and mechanical action by stirring
water; Griffiths.
(q) Physical constants of gases.
The results of all of these were studied by Rowland in 1880, who himself
tperimented also, and he concluded that the mechanical equivalent of heat
fas nearly
778.6 ft.lbs. = l B.T.U., at latitude of Baltimore,
774.5 ft.lbs. = l B.T.U., at latitude of Manchester.
fith the following corrections to be added for other latitudes.
ititude 10 20 30 40 50 60 70 80 90
ft.lbs 1.62 1.50 1.15. .62 .15 .75 1.41 1.93 2.30 2.43'
Since that time other determinations have been made by Reynolds and
.[orby, using mechanical, and Griffiths, Schuste and Gannon, Callendar and
iarnes, using electrical transformation into heat. Giving these latter deter
linations equal weight with those of Joule and Rowland, the average is
i
1 small calorie at 20 C. (nitrogen thermometer) =4. 181 XlO 7 ergs.
418 ENGINEERING THERMODYNAMICS
On the discussion of these results by Smith, Marks and Davis accept and use
the mean of the results of Reynolds, and Morby and Barnes, which is
1 mean calorie = 4. 1834 X10 7 ergs,
= 3.9683 B.T.U.
1 mean B.T.U. = 777.52 ft.lbs.,
when the gravitational constant is 980.665 cm. sec 2 , which corresponds to 32.174
Ibs., and is the value for latitude between 45 and 46.
For many years it has been most common to use in engineering calculations
the round number 778, and for most problems this round number is still th<
best available figure, but where special accuracy is needed it is likely that nc
closer value can be relied upon than anything between 777.5 and 777.6 for the
above latitude.
' Example. To heat a gallon of water from 60 F. to 200 F. requires the heai
equivalent of how many footpounds?
1 gallon =8.33 Ibs.,
200 F. 60 F. =140 F. rise,
8.33 X 140 = 1665 pounddegrees,
= 1665 B.T.U.
= 778X1665, ft.lbs.
= 90,800 ft. Ibs.
Prob. 1. A feedwater heater is heating 5000 gallons of water per hour froi
40 F. to 200 F. What would be the equivalent energy in horsepower units?
Prob. 2. A pound of each of the following fuels has the heating values as give:
Change them to footpounds.
Average bituminous coal, 14,000 B.T.U. per lb., Average kerosene, 18,000 B.T.U. per l
" small anthracite, 12,000 B.T.U. per lb., " alcohol, 10,000 B.T.U. per
Average gasolene, 20,000 B.T.U. per lb.
Prob. 3. A cubic foot of each of the following gases yields on combusture, t
number of heat units shown. Change them to footpounds.
Natural gas (average), 880 B.T.U. per cu.ft., Carburetted water gas, 700 B.T.U. per cu.
Coal gas, 730 B.T.U. per cu.ft., Mond gas, 150 B.T.U. per cu.ft.
Blast furnace gas, 100 B.T.U. per cu.ft.
Prob. 4. A pool contains 20,000 cu.ft. of water and must be warmed from 40
to 70 F. How much work might be done with the equivalent energy?
HEAT AND MATTER 419
Prob. 6. How many calories and how many centigrade heat units would be
required in Prob. 4?
Prob. 6. In the course of a test a man weighing 200 Ibs. goes up a ladder 25 ft.
high, every 15 minutes. If the test lasted 12 hours how many B.T.U. did he expend?
Prob. 7. A reservoir contains 300 billion gallons of water which are heated each
year from 39 F. to 70 F. What is the number of footpounds of work equivalent?
Prob. 8. A pound of water moving at the rate of 450 ft. per second is brought
to rest, so that all of its energy is turned into heat. What will be the temperature rise?
Prob. 9. For driving an automobile 30 horsepower is being used. How long
will a gallon of average gasolene, sp.gr. = .7, last, if 10% of its energy is converted
into work?
Prob. 10. Power is being absorbed by a brake on the flywheel of an engine.
If the engine is developing 50 horsepower how many B.T.U. per minute must be
carried off to prevent burning of the brake?
5. Temperature Change Relation to Amount of Heat, for Solids, Liquids,
Gases, and Vapors, not Changing State. Specific Heats. Provided gases do not
decompose, vapors condense, liquids freeze or evaporate, and solids melt, under
addition or abstraction of heat, there will always be the same sort of relation
between the quantity of heat gained or lost and the temperature change
for all, differing only in degree. As the reception of heat in each case
causes a temperature rise proportional to it and to the weight of the sub
stances, this constant of proportionality once determined will give numerical
relations between any temperature change and the corresponding amount of
heat. Making the weight of the substance unity, which is equivalent to the
consideration of one pound of substance, the constant of proportionality may
be defined as the quantity of heat per degree rise, and as thus defined is the
specific heat of the substance. Accordingly, the quantity of heat for these cases
is equal to the product of specific heat, temperature rise and weight of substance
heated.
The heat, as already explained, may be added in two characteristic ways:
(a) at constant volume or density, or (6) at constant pressure. It might be
expected that by reason of the increase of volume and performance of work
under constant pressure heating, more heat must be added to raise the tempera
ture of one pound, one degree, than in the other case where no such work is done,
and both experimental and thermodynamic investigations confirm this view.
There are, therefore, two specific heats for all substances, capable of definition :
(a) The specific heat at constant volume, and
(6) The specific heat at constant pressure.
These two specific heats are quite different both for gases and for vapors, which
suffer considerable expansion under constant pressure heating, but for solids
and liquids, which expand very little, the difference is very small and is to be
neglected here. As a matter of fact, there are no cases of common engineering
practice involving the specific heat of liquids and solids under constant volume,
and values for the specific heats of liquids and solids are always without further
definition to be understood as the constant pressure values.
420 ENGINEERING THERMODYNAMICS
Let C, be the specific heat of solids and liquids suffering no change of state.
" C p , be the specific heat of gases and vapors at constant pressure and
suffering no change of state.
" C v , be the specific heat of gases and vapors at constant volume and su
fering no change of state.
" t 2 and ti, be the maximum and minimum temperatures for the process.
" w, be the weight in pounds.
Then will the heat added, be given by.the following equation, if the tempera
ture rise is exactly proportional to the quantity of heat, or in other words,
if the specific heat is constant.
Q = Cw(t2 ti), for solids and liquids . V (603)
Q = C,w(t2 ti), for gases and vapors (not near condensation) when
volume is constant. . (604)
Q = C P w(t 2 1\), for gases and vapors (not near condensation) when
pressure is constant. . . (605)
When, however, the specific heat is variable, as is the case for many sub
stances, probably for all, the above equation cannot be used except when
the specific heat average value, or mean specific heat is used. If the variatior
is irregular this can be found only graphically, but for some substances the
variation is regular and integration will give the mean value. It has been
the custom to relate the specific heat to the temperature above the freez
ingpoint of water, expressing it as the sum of the value at 32 F., anc
some fraction of the temperature above this point to the first and seconc
powers, as in Eq. (606).
Specific heat at temperature (t) =a+&(*32)+c(Z32) 2 . (606
In this equation a is the specific heat at 32, while b and c are constants
different for different substances, c being generally zero for liquids.
When this is true, the heat added is related to the temperature abovf
32 by a differential expression which can be integrated between limits
Q = C* * 2 [a+b(t32)+C(t32) 2 ]dt
Jti 32
Q9"\3 ft 39^^31 (&(\7
OA) (li OA) J. ^OU<
Usually the heats are calculated above 32 so that the heats between an
two temperatures will be the difference between the heats from 32 to thos
two temperatures. In this case i = 32, and, t2 = t, whence
B.T.U. per lb., from 32 to *,= U+(*32)+(*32) 2 ](*23). . .
(608
HEAT AND MATTER 421
For this range of temperature 32 to t, the quantity of heat may be ex
Dressed as the product of a mean specific heat and the temperature range
)r
Heat from 32 to t = (mean sp. heat from 32 to t) X (t 32). . (609)
Comparing Eq. (608) with Eq. (609) it follows that
( Mean specific heat 1 ,&/, OON.C/, o 9 x 2 /f. 1n x
. Q9ir +^ /o ^ ) ato^o^r^,^ (oio)
jfrom 32F.to*F.J ' 2 V " ^3
The coefficient of (32) in the mean specific heat expression, is half that in
j,he expression for specific heat at t, and the coefficient of {t 32) 2 , is onethird.
Fhis makes it easy to change from specific heat at a given temperature
ibove 32, to the mean specific heat from 32 to the temperature in question.
The specific heats of some substances are directly measured, but for some
)thers, notably the gases, this is too difficult or rather more difficult than cal
;ulation of values from other physical constants to which they are related.
 It often happens that in engineering work the solution of a practical
problem requires a specific heat for which no value is available, in which case
he general law of specific heats, known as the law of Dulong and Petit, for
ilefinite compounds may be used as given in Eq. (611).
(Specific heat of solids) X (atomic weight) = 6.4. . . . (611)
This is equivalent to saying that all atoms have the same capacity for
seat, and while it is known to be not strictly true, it is a useful relation in
lie absence of direct determinations. Some values, experimentally determined
or the specific heats of solids, are given in Table XXXI at the end of this Chap
er, together with values calculated from the atomic weights to show the degree
f agreement. The atomic weights used are those of the International Com
aittee on Atomic Weights (Jour. Am. Chem. Soc., 1910). When the specific
eat of a solid varies with temperature and several determinations are avail
ble, only the maximum and minimum are given with the corresponding tem
eratures, as these usually suffice for engineering work.
To illustrate this variability of specific heat of solids, the values deter
lined for two samples of iron are given in Figs. 120 and 121, the former
bowing the variation of the mean specific heat as determined by Oberhoffer
nd Harker from 500 F. up, and the latter the amount of heat per pound of
xm at any temperature above the heat content at 500 F., which is gen
ially called its total heat above the base temperature, here 500 F.
; It is extremely probable that the specific heats of liquids all vary irregularly
tilth temperature so that the constant values given in Table XXXII at the end
f the Chapter must be used with caution. This is certainly the case for water,
Id is the cause of the difficulty in fixing the unit of heat, which is best solved
!y the method of means. In Fig. 122 are shown in curve form the values for the
422
ENGINEERING THERMODYNAMICS
specific heats of water at temperatures from 20 F. to 600 F., as accepted
by Marks and Davis after a critical study of the experimental results of
.18
(a)
(M
Obert
Hark
offer
r
/
^
X
_
(a)
Jl6
tt
//
sU
(!>)
Mean Specific He
x ^
/
x^
,'
x
x x
.10
f
1(
00
^Temperature in Degrees Fahr.
FIG. 120. Mean Specific Heat of Iron above 500 F., Illustrating Irregular Variations not
Yielding to Algebraic Expression.
Barnes and Dieterici and adjustment of the differences. The integral curve
is plotted in Fig. 123 which, therefore, gives the heat of water from 32F. to any
/
'(a)
B.T.Us.Per Pound of Iron Above 500]
8 1 1
(a)
W
From
Oberhof fer D
Harker
ata
s
/
s
^
\, '.
/ s
^'
S
,S
<y
S
7*
'"
s'
500
1000
2500
1500 2000
Temperature in Degrees Fahr .
FIG 121. Total Heat of Iron above 500 F., Illustrating its Approximation to a Straigl
Line Relation in Spite of Wide Variation in Specific Heat Given in Fig. 120.
temperature up to the highest used in steam practice and which is designate
in steam tables, summarizing all the properties of water and steam, as tl
HEAT AND MATTER
423
100
200 300 400
Temperature Degrees Fahr.
500
FIG. 122. Specific Heat of Water at Various Temperatures.
/
/
200 400
Temperature in Degrees Fahr.
600
IG. 123. Total Heat of Water from 32 F., to any Temperature, the Heat of the Liquid at
that Temperature above 32 F.
424
ENGINEERING THERMODYNAMICS
eat
heat of the liquid. For the purpose of comparison, the mean specific h
of water is given in Fig. 124 from 32 F. to any temperature which is obtained
from the heat of the liquid above 32 F. by dividing it by the temperature
above 32 F.
In the table of specific heats of liquids there is a column giving the value
calculated from the atomic weights to show at a glance the degree with
which liquids satisfy the Dulong and Petit law.
Variability of specific heat is especially noticeable in liquids that are solu
tions with different amounts of dissolved substance, in which case the specific
heat varies with the density and temperature. Problems of refrigeration
involve four cases of this kind: (a), calcium, and (6), sodium chloride,
L06
1.04
1.0?
LOO
200 400 600
Temperature in Degrees Fahrenheit
FIG. 124. Mean Specific Heat of Water from 32 to any Temperature.
brines, the densities of which vary considerably but which are used with
but little temperature range, seldom over 20 F. and often not over 5 F. ?
(c), anhydrous ammonia and (d), carbonic acid.
As the density of brines is often reported on the Baume scale and liquid
fuels always so, a comparison of this with specific gravities is given in Table
XXXIII in connection with the specific heat tables at the end of this Chapter
to facilitate calculation.
One of the bestknown solutions so far as accuracy of direct experimental
data is concerned, is calcium brine, results for which, from 35 C. to 20 C.
given below, are from U. S. Bureau of Standards Bulletin by Dickinson,
Mueller and George, for densities from 1.175 to 1.250. For chemically pure
HEAT AND MATTER
425
calcium chloride in water, it was found that the following relation be
tween density D, and specific heat C, at C.,
D = 2.8821  3.6272C+ 1.7794C 2 ,
(612)
and these results plotted in Fig. 125 show the specific heat variation with
temperature to follow the straight line law very nearly. This being the case
the mean specific heat for a given temperature range is closely enough the
arithmetical mean of the specific heat at the two limiting temperatures. To
fthe figure are also added dotted, the specific heats for some commercial brines,
not pure calcium chloride, but carrying magnesium and sodium chloride of
density 1.2.
It might be conveniently noted here that the relation between freezing
point and density for pure calcium chloride by the same bulletin is given
in Table XVIII below:
TABLE XVIII
FREEZINGPOINT OF CALCIUM CHLORIDE
U. S. BUREAU OF STANDARDS
Density of Solution.
Per cent CaClt by Wt.
Freezingpoint,
C.
Freezingpoint,
1.12
14.88
 9
15.8
1.14
16.97
13
8.6
1.16
19.07
16
3.2
1.18
21.13
20
 4.0
1.20
23.03
24
11.2
1.22
24.89
29
20.2
1.24
26.77
34
29.2
1.26
 28.55
40
40.0
Other values for the specific heats of brines as commonly used are given
in Table XIX, the accuracy of which is seriously' in doubt and which
may be checked by more authoritative values at different points where deter
minations have been made.
Anhydrous ammonia liquid, has a variable specific heat with temperature,
.put the experimental values are too few to make its value and law quite certain,
lieveral formulas have been proposed, however, that tend to give an impression
(of accuracy not warranted by the facts though quite convenient in preparing
tables.
Authority Specific heat of NHi liquid at t" F.
Zeuner 1.0135+.00468 032) (a)
Dieterici 1.118 +.001156 032) (6)
Wood 1.1352+.00438 032) (c)
Ledoux 1.0057 +.00203 032) (d)
(613)
426
ENGINEERING THERMODYNAMICS
I
.75
I
.70
.GO
10
D=l
D=l.
=1.22,
70
10 30 50
Temperature in Degrees Fahr.
FIG. 125. Specific Heat of Calcium Chloride Brine of Various Densities D at Temperatures
10 F. to +70 F.
HEAT AND MATTER
427
TABLE XIX
SPECIFIC HEAT OF SODIUM CHLORIDE BRINE
Density, B6
Sp.gr.
Per cent NaCl
by Wt.
Sp. Heat.
Temp. F.
Authority.
1
1.007
1
1.6
.992
978
0
64 4
Common
Thomsen
4 9
995
66 115
5
10
1.037
1.073
5.0
10.0
10.3
.960
.892
.892
0
0
59 120
Common
Common
Teudt
10 3
912
59 194
Teudt
11.5
.887
61126
Marianne
12 3
871
64 4
\Vink elm a,nn
15
1.115
15.0
18 8
.892
841
0
63 125
Common
Teudt
18.8
.854
68  192
Teudt
19
1.150
20.0
24.3
.829
.7916
6468
Common
Winkelmann
24.5
.791
64
Thomsen
23
1.191
25
.783
Common
From these expressions the mean specific heat follows by halving the coefficient
of (t 32) F., and these were determined and plotted to scale, together with some
direct experimental values of Drewes, in Fig. 126. Giving greatest weight
[to Drewes and Dieterici, a mean curve shown by the solid line is located as
the best probability of the value for liquid anhydrous and it has the Eq. (614).
Mean specific heat of anhydrous 1 _i rt7
liquid NH 3 from 32 F.toW.
32). . . . (614)
From this value the heat of liquid ammonia above 32 F. has been determined
and is presented graphically in Fig. 127 from which, and the equation, the
tabular values at the end of the Chapter were determined.
Ammonia dissolved in water, giving an aqueous solution as used in the absorp
tion refrigerating system, has a nearly constant specific heat so closely approxi
mating unity as shown by Thomsen, who gives
3 per cent NH 3 in water solution.^?.^, = .997, at 66 F.
1.8 per cent NH 3 in water solution, sp.ht., =.999, at 66 F.
.9 per cent NH 3 in water solution, sp.ht., =.999, at 66 F.,
Ithat it is customary in these calculations to ignore any departure from unity,
the value for water.
Liquid carbonic acid, another important substance in engineering, especially
pn mechanical refrigeration, is less known as to its specific heat than is ammonia,
land that is much too uncertain. There is probably nothing better available
at present for the necessary range than the results of Amagat and Mollier,
reported by Zeuner for the heat of the liquid, which are reproduced in Fig.
128, and used in the table at the end of this Chapter.
428
ENGINEERING THERMODYNAMICS
It is, however, with gases that the most complex situation exists with respect
to specific heats. As has already been pointed out, gases may be heated at
1.4
I 1  2
OQ
I"
50
"A
AAZeuner
BBLedoux
CCWood
DD Dietetic!
E.E Drewes
FFMean Used in Book
+ 50
Temperature in Degrees Fahr.
150
FIG. 126. Mean Specific Heat of Liquid Anhydrous Ammonia from 50 F. to 150 .F
I
t3
S
i
wu
100
D
E
/
/
^
/
/
/
/
/
/
,
/
/
^
Cx
,
^
^
A
0g
X^
AA Zeuner
B Wood Note: All Curves
C Dieterici Practically Coin
D Ledoux cident above 32
E Mean
**
A
^
^
^
^
^
^
ivi: 50 50 100 150 200
Temperature in Degrees Fahr.
FIG. 127. Heat of Liquid Anhydrous Ammonia above 50 F.
constant volume, doing no external work while being heated, or at constant
pressure, in which latter case work is done by expansion of the gas against the
resisting constant pressure. Therefore, there must be two different specific
HEAT AND MATTER
429
heats for each gas, one C v at constant pressure and the other C c at constant
volume, the difference between them representing the heat equivalent of the
50
25
I
i_l
M
O
3
25
7
7
25 25 <o
Temperature in Degrees Fahr.
FIG. 128. Heat of Liquid Carbonic Acid above 32 F.
)rk of expansion done during the rise of temperature. Most experimental
leterminations of the specific heats of gases have been made at constant pressure
430 ENGINEERING THERMODYNAMICS
and the constant volume value found from established relations between it
and other physical constants. These relations most commonly used are two,
Eq. (615) connecting the difference with a constant R and the other Eq. (616)
777.52(C P C,) = R,
^ = Y (616)
Op
/
connecting their ratio to a constant y. These constants have each
a special significance that may be noted here and proved later, thus R is the
ratio of the PF product of a pound of gas to the absolute temperature, and y the
particular value taken by the general exponent s in PV s = c, when the expansion
represented takes place with no heat addition or abstraction, i.e., adiabatic,
it is also a function of the velocity of sound in gases. Table XXXIV at the end
of this Chapter gives some authentic values, with those adopted here designated
by heavy type.
Variability of specific heats of gases and vapors is most marked and of some
engineering importance, because so many problems of practice involve highly
heated gases and vapors, the most common being superheated steam and the
active gases of combustion in furnaces, gas producers and explosive gas engines,
In fact, with regard to the latter it may be regarded as quite impossible witrJ
even a fair degree of accuracy to predict the temperature that will result in thei
gaseous products from the liberation of a given amount of heat of combustio
The first fairly creditable results on the variability of the specific heats of gas
of combustion at high temperatures were announced by Mallard and LeChatelie
Vieille and Berthelot, all of whom agree that the specific heat rises, but wh
do not agree as to the amount. A general law was proposed by LeChatelie
giving the specific heat as a function of temperature by an equation of tl
following form:
Specific heat at * F., (F=C),=C, = a+6(*32), ._ (61
in which a = specific heat at constant volume at 32 F. This yields,
jB.T.U.perlb.from32 )_ n ^
32) ..... (61
J Mean specific heat from 1 , 6 ,
The specific heat at constant pressure is obtained by adding a consta
to the value for constant volume according to
whence
Specific heat at t F.,
HEAT AND MATTER
~Y\ /"f t\>
431
(621)
S^. . (622)
Mean specific heat from 1 _
32 F., toZF., (P = C) ~
_ r / _ .
~ Up ~ " t
6.
, .
(b }
The values of these constants have been determined by LeChatelier, Clerk,
I Callender, and Holborn and Austin, from which the following values are
I selected.
TABLE XX
SPECIFIC HEAT CONSTANTS, GASES,
Gas.
a
R
+777T52
6
b
2
Authority.
CO 2
C0 2
N 2
N 2
N 2
.1477
.170
.1944
.2010
.2404
.2350
2350
.000097
.0000824
.0000484
.000021
0000208
.0000484
.0000412
.0000242
.0000105
0000104
LeChatelier
Holborn and Austin
LeChatelier
Holborn and Austin to 2606 F.
Callender 1544 F to 2440 F
2
H 2 O
1 Air
.1488
.3211
.2125
.2431
.0000424
.000122
.000135
.0000212
.000061
. 0000675
LeChatelier
LeChatelier
Callender (1544 F. to 2440 F.)
For purposes of comparison the following curves are plotted, showing all
these results of specific heat at constant volume, at temperature t F., the total
heat above 32 F. per pound of gas, and the mean specific heat from 32 F. to
t F. in Fig. 129.
Probably there is now more known of the specific heat of superheated steam
than of any common gaseous substance, and it is likely that other substances
will be found in time to have somewhat similar characteristics. Pure computa
tion from the laws of perfect gases indicates that the specific heat of gases or
 superheated vapors must be either a constant, or a function of temperature
s only, and this is what prompted the form of the LeChatelier formula. Bold
I experimentation on steam, disregarding the law, or rather appreciating that
superheated steam is far from a perfect gas, principally by Knobloch and
f Jacob and by Thomas, showed its specific heat to be a function of both pres
sure and temperature. Results were obtained that permitted the direct solu
iftion of problems of heat of superheat, or the heat per pound of vapor at any
[temperature above that at which it was produced, or could exist in contact with
jjthe liquid from which it came. Critical study of various results by Marks and
4 Davis led them to adopt the values of Knobloch and Jacob with slight modifi
1
432
ENGINEERING THERMODYNAMICS
^
\
\
\r\
\
\\
1 J
5 03 "
2 fe
> S <N
!o> w
'1 I
1
*2 '~
~ %
I
o
\
N^N
s.
\ .
\
\
V
a
'i
. 1
^
> Q J
J 1
Curve 7 Holborn & Austin's Values for COo
 8  " N/
^^
*\
V^
\
\
V
\
\
\ s
\
\
\
N
\
N^
, s
\
N^
\
\
\
Curves 16 from values of
LeChatelier and Clerk
Solid Lines = Sp.Ht. at Cons
Broken"
1 ftn.l 1 POn 9. nnrl S  Vn
'
\
\
iV
\
\ s
s
\
\
X
\
y \
\
\
N
^
>,
\
\
S. N
*
\N
\
S
\
\
\
\
\
\ *
\
^4 N
\
V
\
^ \
\
s\
\
N
\ s
\.
HEAT AND MATTER
433
I
m
fl'X'lI.uoi^jn^s 9AOQB UIROJS jo puno c t aoj ^OH
oanssojj ^UB^SUOO ^1 IBOH oO'^^tlS a3W
434 ENGINEERING THERMODYNAMICS
cations, for which evidence was in existence, raising the specific heats at low
pressures and temperatures, and their conclusions are adopted in this work.
In Fig. 130 is shown (A) the Marks and Davis modification of the C p curve
*of Knobloch and Jacobs, the integral of which (C) gives the heat of superheat
from any temperature of steam generation to actual steam temperature, while
(B) shows the values for the mean specific heat above the temperature of satura
tion for the particular pressure in question.
When substances of the same class are mixed so that Wi, W2, w^, etc., Ibs.
of the different substances having specific heats Ci, C 2) 3, etc., or C P i, C P 2,
C P 3, etc., or Cci, ,2, C P s, etc., then the specific heat of the mixture is given by
f .
Example. If 5 Ibs. of olive oil at a temperature of 100 F., 10 Ibs. of petroleum
at a temperature of 150 F., and 50 Ibs. of water at 50 F. are mixed together, what
will be the resultant temperature and how much heat will be required to heat the mix
ture 100 above this temperature?
Sp. ht. of olive oil =* .4,
Sp. ht. petroleum = .511,
Sp. ht. water =1.000.
Let z=the final temp. The heat given up by the substances falling in tempera
ture is equal to that gained by those rising, hence
50(x50) XI = 5(100 a;) X. 4 +10(150 a;) X.511,
50x 2500 =200 2x +766 5.113,
57.11z =3466, or, y, =60.7 F.,
Sp.ht. of mixture = 1 " /1 ^ c ^" r ^ 3 > f rom Eq (611)
Wi \W 2 +1#3
5X.4+10X.511+50X1 _57.11
5+10+50 = 65
whence the heat required will be 65 X. 8786 = 57 B.T.U.
Prob. 1. To change a pound of water at 32 F. to steam at 212 F. requires
1150.4 B.T.U's. If the same amount of heat be given to a cubic foot of each of
the following substances at 32 F., what will be final temperature in each case? (a) cop
per; (6) iron; (c) mercury; (d) clay; (e) stone.
HEAT AND MATTER 435
Prob. 2. How many pounds of the following substances could be warmed 10 F.
y the heat required to raise 100 Ibs. of water from 40 F. to 200 F.?
(a) Ethyl alcohol from 100 F.;
(6) Sea water from 60 F., (density = 1.045);
(c) Glycerine from 60 F;
(d) Tin from 480 F.
Prob. 3. If 150 Ibs. of water at 200 F. are added to a tank containing 200 Ibs
.f petroleum at 70 F., what will be the resultant temperature, neglecting any heat
Absorbed or given up by the tank itself?
Prob. 4. To melt 1 Ib. of ice requires 144 B.T.U. How much would this lower
he temperature of 1 Ib. of the following substances (1) at constant pressure; (2) at
sonstant volume; (a) air; (6) oxygen; (c) ammonia; (d) hydrogen; (e) nitrogen?
Prob. 5. What would be the specific heats of the following mixture? Hydrogen
\ Ibs., oxygen 1 Ib., nitrogen 7 Ibs., carbon dioxide 20 Ibs., carbon monoxide 10 Ibs.?
Prob. 6. Air is approximately 77 per cent N2, and 23 per cent 02 by weight. By
paeans of the specific heats of the components, find its specific heats at constant pres
jure, and at constant volume.
Prob. 7. By means of the specific heats, find the values of R and y most correct
tit atmospheric temperature (60 F.) for, hydrogen, air, carbon dioxide, carbon monoxide
a,nd nitrogen.
Prob. 8. How much water could be heated from 40 F. to 60 F. by the heat
needed to superheat 10 Ibs. of steam at 200 Ibs. per square inch absolute to 700 F.?
Prob. 9. A building containing 250,000 cu. ft. of space is heated by a hotwater
:3ystem. Considering the air to change eight times per hour, how many pounds of
prater per hour must be circulated if the drop in temperature of the water is from
J200 to 100 and the temperature of the outside air is 30 F. while that of the room
jis 60 F. neglecting wall conducted heat?
Prob. 10. How much heat would be required to warm a pound of liquid C0 2 from
(zero to 80 F.? Compare with water and ammonia.
6. Volume or Density Variation with Temperature of Solids, Liquids, Gases
(and Vapors, Not Changing State. Coefficients of Expansion. Coefficients
j of Pressure Change for Gases and Vapors. Solids increase in length or in any
1 linear dimension, a certain fraction of their original length for each degree
temperature rise and the expansion is usually assumed to be in proportion
to temperature rise. The relation between original and final length can be
[set down in an equation involving the coefficient of expansion.
Let a = coefficient of linear expansion = fractional increase in length per
I degree.
" h and ti = original length or any other linear dimension and the cor
responding temperature;
and t2 = length which h becomes after heating and the corresponding
temperature.
Increase in length = h Ii = ali(t2 ti), (627)
New length Z 2 = Zi+aZife h),
fe*i)] (628)
436 ENGINEERING THERMODYNAMICS
Solids, of course, expand cubically and the new volume will be to the old
as the cubes of the linear dimension.
Let a = coefficient of volumetric expansion;
vi = original volume;
V2 = final volume after heating.
Then when the temperature rises one degree,
= l+a .... (629
If a is small, and it is generally less than ioooo> then the square and cube cai
be neglected in comparison with the first power, whence
l+a = l+3a and a = 3a.
so that the coefficient of volumetric expansion may be taken as sensibb
equal to three times the coefficient of linear expansion, and similarly, th>
coefficient of surface expansion as twice the coefficient of linear expansion.
Liquids, by reason of the fact that they must always be held in solu
containers, may be said to have no linear expansion, and therefore, althougl
the expansion may be one direction only, the amount is due to the tota
change of volume rather than the change of length along the direction o
freedom to expand. The same is true of gases, so that for gases and liquids
only coefficients of volumetric expansion are of value and these are given in th
Tables XXXV, XXXVI, XXXVII, and XXXVIII at the end of this Chapter
With liquids and gases it is usual to take the volume at C. or 32 F. anc
29.92 ins. Hg pressure as a standard, and the coefficient gives the increase a:<
a fraction of this, per degree departure from the freezingpoint. This is tht
universal practice with gases.
It appears that the coefficients of expansion for solids are quite differen
from one another, ranging from over 15X10" 4 for wax, to .085X10 ~ 4 fo
Jena normal glass, a range of over two hundred and sixty times. Determina
tions of the value at various temperatures for any one substance indicate i
variation with temperature, which proves that proportionality of increase o
dimensions to temperature rise, does not hold true, a. fact which has led t<
formulas of the form
the value of which is dependent on the determination of the constant and veri
fication of correctness of form, which has not by any means been conclusively
done. For most engineering work the constant values nearest the temperature
range will suffice except for certain liquids, vapors, and gases. A more markec
tendency to follow such a law of variation with temperature is found witl
liquids and coefficients for some are given in the standard physical tables.
HEAT AND MATTER 437
The two important liquids, mercury and water, have been separately
studied in greater detail and the latter exhibits a most important exception
to the rule. For mercury, according to Broch
which exhibits a refinement of value only in instrument work such as barometers
ind thermometers. Water, as already mentioned, has its maximum density
at 39.1 F. and expands with both fall and rise of temperature. Its expansion
jb given by a similar formula by Scheel, as follows:
. . (631)
VTost commonly the expansion of water is not considered in this'way, but by
omparing densities at varying temperatures, and all sets of physical tables
ontain values which in this work are significant only as affecting the change
f volume in turning water to steam and such values as are needed are
ncorporated in the steam tables later.
The study of the expansion of gases and vapors at constant pressure, and
ise of pressure at constant volume, per degree has perhaps been fairly com
lete and is of greatest significance, because from it most of the important laws
f thermodynamics have been derived. This work may be said to have
tarted with the Regnault air and gas thermometer work, already described,
ome of the authentic values collected in the Landolt, Bornstein, Myerhoffer,
ind Smithsonian Physical Tables, are given at the end of this Chapter,
here a p is the coefficient of pressure change at constant volume, and <x p the
coefficient of expansion, or volume change at constant pressure.
The remarkable thing about the coefficients for these gases and vapors is the
pproach to constancy for most of the gases, not only of the coefficients of expansion
or P = c nor the similar constancy of the coefficients of pressure rise for V = c, but
wre remarkable than either of these is the similarity of the two constant coeffi
lents. These facts permit of the generalizing of effect when P = c,
d when V = c, and of the announcement of a law by means of which
fl such problems can be solved instead of applying separate coefficients for
very substance and every different temperature necessary for solids and
quids where, for example, the maximum coefficient was over 260 times as
reat as the least. The average coefficient for all gases, applying both to
pressures and volumes, is the same as enters into the gas thermometer work
lid its best value is found to be
a =  = .002034, per degree F. '
, . . , . (632)
a = 27^3 = 003661 , per degree C.
438 ENGINEERING THERMODYNAMICS
and approximately
a = J = .00203, per degree F.
a = J =.00366, per degree C. ,
i o
. (633)
These are the same as the reciprocals of the absolute temperature of the
icemelting point, and are but expressions of conditions for reduction of the
volume and pressure at the icemelting temperature to zero by constanl
pressure and constant volume abstraction of heat respectively, and bj
stating the amount of reduction per degree give by implication the numbei
of degrees for complete reduction. J
Example. The rails on a stretch of railroad are laid so that they just touch whei
the temperature is 120 F. How much total space will there be between the rallj
per mile of track at F.?
For wrought iron a will be nearly the same for Bessemer steel = .00000648.
Hence the linear reduction in 5280 ft. for a change of 120 F. will be
5280 X 120 X .00000648 = 4. 1 f t.
Prob. 1. A steam pipe is 700 ft. long when cold (60 F.), and is anchored at or
end. How much will the other end move, if steam at a temperature of 560 F.
turned into the pipe?
Prob. 2. A copper sphere is one foot in diameter at 50 F. What must be tl
diameter of a ring through which it will pass at a temperature of 1000 F.?
Prob. 3. A hollow glass sphere is completely filled with mercury at F. Whr
per cent of the mercury will be forced out if the temperature rises to 300 F.?
Prob. 4. A room 100 ft.xSO ft.XlO ft. is at a temperature of 40 F. The teni
perature rises to 70 F. How many cubic feet of air have been forced from the roon
Prob. 5. The air in a pneumatic tire is at a pressure of 90 Ibs. per square inn
gage and at a temperature of 50 F. Due to friction of the tire on the ground
running, the temperat re rises to 110 F. What will be the pressure?
Prob. 6. A brick lighthouse is approximate y 200 ft. high. Should it be exact
this at F., what would it be at 100 F.?
Prob. 7. Show that if a glass tube is rigidly held at each end by brackets attach
to an iron tank it will break if the tank is warmed.
Prob. 8. From Eq. (618) find the density of water at 60 F., 100 F., 21
F., and compare with the values in the steam tables.
Prob. 9. A drum containing C0 2 gas at a pressure of 250 Ibs. per square in
gage is raised 100 F. above its original temperature. What will be the new pressui
7. Pressure, Volume and Temperature Relations for Gases. Perfect a
Real Gases. Formulating the relations between the pressure change at consts
volume and the volume change at constant pressure,
Let P and V be the simultaneous pressure and volume* of gas;
t be its scale temperature at the same time, F. ;
" T be its absolute temperature at the same time, F.
HEAT AND MATTER 439
Chen at constant volume the pressure reached at condition (a) after heating
rom 32 F. is given by

Similarly for another temperature t b , the pressure will be
Vhence
492
1 ,
"
492
~ = ^ } for V constant, (634)
T& It,
Similarly
^ = 1^ for P constant (635)
V 6 J ft
jioth Eqs. (634) and (635) are true, for no gas all the time, but very nearly
[rue for all, under any range of change, and a hypothetical gas is created for which
It is exactly true all the time, known as a perfect gas, about which calculations
ian be made as would be impossible for real gases and yet the results of which
Ire so close to what would be the result with real gases, as to be good enough
jor engineering practice. Therefore, with a mental reservation as a guard
Igainst too great confidence in the work, all real gases will be assumed perfect
j,nd to follow Eqs. (634) and (635) except when experience shows the results
lire too far wrong to be useful.
These laws, known by the names both of Charles and GayLussac, are closely
jissociated with another also doubly named as Boyle's or Mariotte's and like
Use an idealization of experimental observations known to be nearly true for
I) 11 gases. This is to the effect that so long as temperatures are kept constant
he pressures of gases vary inversely as their volume, or that,
=*, and, P Q V a = PbV b = constant, for T constant . (636)
* b V a
Study of the PV product, for various gases has revealed a good deal on the
leneral properties of matter, especially as to the transition from one state to
mother. This is most clearly shown by curves which may be plotted in two
440
ENGINEERING THERMODYNAMICS
HEAT AND MATTEK 441
ways. To coordinates of pressure and volume a family of equilateral hyper
bolas one for each temperature, would represent the true PV = C or isothermal
relation and any variation in the constancy of the product would be shown by
its departure from the hyperbola. Still more clearly, however, will the depart
ure appear when the product PV is plotted against pressures, constancy of
product would require all lines to be straight and inconstancy appear by
departures from straight lines. To illustrate, the data from Young for car
bon dioxide are plotted both ways in Fig. 131, from 32 F. to 496 F., the values
of PV at 32 and 1 atm. are taken as unity on one scale. It appears that up to
the temperature of 88 F, known as the critical temperature, each isothermal
plotted to P and PV coordinates consists of three distinct parts:
(a) a curved line sloping to the right and upwards;
(6) a straight line nearly or exactly horizontal;
(c) a nearly straight line sloping upward rapidly and to the left.
In this region then the isothermals are discontinuous, and this is caused by
the liquification or condensation of the gas, during which increase of pressure,
produces no change of volume, provided the temperature is low enough. It
also appears that each PV line has a minimum point and these minima joined
result in a parabola. At the end of this Chapter are given in Table XXXIX the
values of PV at three different temperatures and various pressures for oxygen,
hydrogen, carbon dioxide and ammonia, in terms of the values at 32 and 1
atm. for further comparison and use. Further study along these lines is not
profitable here and the topic while extremely interesting must be dropped with
the observation, that except near the point of condensation or liquefaction,
gases or vapors, which are the same thing except as to nearness to the critical
state, follow the Boyle law closely enough for engineering purposes.
None of these approximate laws, Eqs. (634), (635) and (636) can be con
sidered as general, because each assumes one of the variables to be constant, but a
general law inclusive of both of these follows from further investigation of a
fixed mass of gas suffering all sorts of pressure volume and temperature changes,
such as occur in the cylinders of compressors and gas engines. A table of
simultaneous experimental values of pressure, volume, and temperature, for any
gas will reveal the still more general relation inclusive of the preceding three as
follows :
P.V. P>V b _PV_ ,
Iv "IT T "
iii which C g is approximately constant for any one gas and assumed constant
for perfect gases in all calculations. For twice the weight of gas at the same
pressures and temperatures C ff would be twice as large, so that taking a constant
R for one pound, and generally known as the " gas constant," and introducing
a weight factor w, the general characteristic equation for the perfect gas is,
(638)
442
ENGINEERING THERMODYNAMICS
This general law may be derived from the three primary laws by imagining
in Fig. 132, two points, A and B, in any position and representing any two states
of the gas. Such points can always be joined by three lines, one constant
Diagram to derive Law ^=C^
FIG. 132. Curve of Continuous Relation between P, V, and T for Gases.
pressure A to X, one constant temperature X to Y, and the other constant
volume Y to B. For these the following relations hold, passing from A to B
T
V V
V a V xm
But
and
whence
P
V V v.
V X V yp ,
P T
V T7 v a
v a V vr> 7fT>
or
Passing to B,
?sZs = Vz
T. T,
T,
'TV
HEAT AND MATTER 443
But
or in general
T
= fM
Ty Ty
PV
T = constant. = wR
when the weight of gas is w Ibs.
For numerical work, the values of R must be fixed experimentally by direct
measurement of simultaneous pressure, volume, and temperature, of a known
weight of gas or computed from other constants through established relations.
One such relation already mentioned but not proved is
R = 777.52(C P C V ). . ....... (639)
It is extremely unlikely that the values of R found in both ways by a multi
tude of observers under all sorts of conditions should agree, and they do not,
but it is necessary for computation work that a reasonable consistency be attained
and that judgment in use be cultivated in applying inconsistent data. In the
latter connection the general rule is to use that value which was determined by
measurement of quantities most closely related to the one being dealt with.
Thus, if R is to be used to find the state of a gas as to pressure, volume, and tem
perature, that value of R determined from the first method should be selected, but
the second when specific heats or Joule's equivalent are involved. Of course, a
consistency could be incorporated for a perfect gas, but engineers deal with real
gases and must be on guard against false results obtained by too many hypoth
eses or generalizations contrary to the facts. Accordingly, two values of R are
given in Table XL, at the end of this chapter, one obtained from measure
ments of specific heats at constant pressure and determinations of the ratio
of specific heats unfortunately not always at the same temperature and gen
erally by different people, and the other by direct measure of gas volume at
standard 32 F. temperature and 1 atm. pressure. These measurements are
separately reported in Sections (5) and (8), respectively.
If a gas in condition A, Fig. 133, expand in any way to condition B, then
it has been shown that
in which /has any value and which becomes numerically fixed only when the
process and substance are more definitely defined. Comparing the temperatures
at any two points A and B, it follows that
444
whence
But
and
whence
ENGINEERING THERMODYNAMICS
T a PaVa
PaVa \V
~
and
T a \V b J
A
\
\
\
\
X
\
^^
B
V
don or Compression of Gas between A and B, Causing a Change c
perature.
Vb /7V\rH
V~~ V 7\ /
(640)
(641)
Eqs. (640) and (641) give the relation between temperatures and volume?
But
=
V, P a T b
which, substituted in above, gives
T b P
or
HEAT AND MATTER 445
and
o \ a
:>r
irHir , (642)
J. a \JTn /
(643)
Eqs. (629) and (630), give the relation between pressures and temperatures.
It is convenient to set down the volume and pressure relations again to
complete the set of three pairs of most important gas equations.
(644)
(645)
iese are perfectly general for any expansion or compression of any gas, but
we of value in calculations only when s is fixed either by the gas itself or by
;he thermal process as will be seen later.
Example. A pound of air has a volume of 7.064 cu.ft. at a pressure of two atmos.
)heres and a temperature of 100 F. Find the value of R for air from the data;
,lso the final volume and temperature if expansion occurs so that s=*1.4 until the
ressure becomes ^ an atmosphere.
PV=wRT, or 2116x2x7.064 = 1X^X560, or #=53.38,
sl .4
.. !F 2 = 7^1.49= =352 abs. = 108 F.
1.49
i
? = ^V =2.7, or 7t2.77i19.lcu.ft.
I Prob. 1. A perfect gas is heated in such a way that the pressure is held constant.
r the original volume was 10 cu.ft, and the temperature rose from 100 F. to 400
j., what was the new volume? t ,
I Prob. 2. The above gas was under a pressure of 100 Ibs. per square inch gage at
je beginning of the heating. If the volume had been held constant what would have
iaen the pressure rise?
 Prob. 3. A quantity of air, 5 Ibs. in weight, was found to have a volume of 50 cu.ft.
id a temperature of 60 F. What was the pressure?
5 Prob. 4. A cylinder holding 12 cu.ft. has a pressure of 250 Ibs. per square
h gage, and the temperature is 50 F. What would be the weight of its contents
ere it filled with (a) C0 2 ; (6) NH 3 ; (c) Oxygen; (d) Hydrogen?
446 ENGINEERING THERMODYNAMICS
Prob. 5. At a pressure of 14.696 Ibs. per square inch and a temperature of
melting ice, one pound of air has a volume of 12.387 cu.ft. From the data find the
value of R for air. The specific heats of air are given by one authority as C p = .2375
and C P = .1685. Find R from the data and see how the two values obtained compare.
Prob. 6. At the same pressure and temperature as in Prob. 5, a pound of the
following substances has a volume as shown. From the data and the values of
specific heats, find R by the two methods.
Substance. Cu.ft. per Ib. C p . C 9 .
Hydrogen 178.93 3.409 2.412
Carbon dioxide.... 8.15 .217 .1535
Oxygen 11.21 .2175 .1551
Nitrogen 12 . 77 . 2438 . 1727
Prob. 7. 5 cu.ft. of gas at a pressure of 3 atmospheres absolute and a tempera
ture of 50 F. expand to atmospheric pressure. What will be the final volume and
temperature, if s = 1.35?
Prob. 8. 1000 cu.ft. of gas at atmospheric pressure and 60 F. are compressec
into a tank of 100 cu.ft. capacity. What will be the pressure in the tank and the
temperature of the gas at the end of the process, if the gas is C0 2 and the com
pression adiabatic?
Prob. 9. What will be the final volume, pressure and temperature, if a pound o:
air at atmospheric pressure (14.7 Ibs. per square inch) and a temperature of 60 F
be compressed adiabatically until its absolute temperature is six times its origina
value?
8. Gas Density and Specific Volume and its Relation to Molecular Weigh
and Gas Constant. The density of a gas is best stated for engineering
purposes as the weight of a cubic foot, but as this becomes less on rise o
temperature or decrease of pressure it is necessary to fix a standard condition
for reporting this important physical constant. It is best to take one atmosphere
760 mm. or 29.92 ins. of mercury as the pressure, and C. = 32F. as the
standard temperature, though it is in some places customary in dealing with
commercial gases, such for example as those used for illumination, to take th
temperature at 60 F. and illuminating gas at this condition is often knowr
among gas men as standard gas. In this work, however, the freezingpoin
and standard atmosphere will be understood where not specifically mentioned
as the conditions for reporting gas density and its reciprocal, the specific volum
of gases or the cubic feet per pound. The chart, Fig. 134, shows the relatioi
of volume and density at any pressure and temperature to the volume am
density under standard conditions.
These constants have been pretty accurately determined by many investi
gators, whose figures, to be sure, do not agree absolutely, as is always th
case in experimental work, but the disagreement is found only in the las
significant figures. Some selected values of reliable origin are reported a
the end of this Chapter in Table XLI for the important gases and
may be used in computation work.
f
HEAT AND MATTER
447
It often happens in dealing with gases and especially superheated vapors
that a value is needed for which no determination is available, so that general
Pressure in Pounds Per Sq. In. Abs.
16 15 14 13 12
10
40 ] 50 00
Temperature , Degrees Fahr.
,80
I I
I I I I
1.2 1.1
I ^P 5 I
i i
I I 156 
I I I I I I I
9 .8
I I
.7
Upper Sea* =Ratlo
Inner Sea,e .
Ou t e, Sca,e 
I I I I I I I I i
.6
Volume at 32 29.92"
Density at 32 F.
Density at any T
FIG. 134. Equivalent Gas Densities At Different Pressures and Temperatures.
, of density or specific volumes of substances are necessary to permit the
led constant to be estimated. These relations may be applied to vapors
448 ENGINEERING THERMODYNAMICS
as well as to gases even though the standard conditions are those for the
liquid state, on the assumption that all gases and vapors will expand under
temperature, or contract under pressure rise, to the same degree, retaining
the same relative relations between all substances as exist at the standard
atmosphere and freezingpoint. A vapor thus reported below its point of
condensation and assumed to have reached that condition from one of higher
temperature at which it exists as vapor is often called steam gas, or alcohol
gas, for example in the case of water and alcohol.
Such general relations between the densities of gases as are so desirable
and useful in practical work have been found by studying the manner in which
gases chemically combine with respect to the volume relations before and after
the reaction. Following several experimenters, who reported observed rela
tions, GayLussac stated a general law, as follows:
When two or more gaseous substances combine to form a compound, the vol
umes of the combining gases bear a simple ratio to each other and also to
that of the compound when it is also a gas.
He also attempted to derive some relation between this law and Dalton's atomic
combining law, which states that, in combining chemically, a simple numerica
relation exists between the number of atoms of different elements which unite
to form a compound. This was not successful, but Avagadro later found tht
expected relation by assuming that it is a particle, or a number of atoms, o
a molecule, that is important in combining, and the law stated is as follows :
Equal volumes of different gases measured at the same pressure and tempera
ture contain the same number of molecules.
It is possible by analysis of these two laws to get a relation between the volume
of gases and the weights of their molecules because the molecular relation o
Avagadro, combines with the combining law of GayLussac to define the rela.
tion between the number of combining molecules. At the same time the weigh
relations in chemical reactions, based on atomic weights, may be put into
similar molecular form, since the weight of any one substance entering is th
product of the number of its molecules present and the weight of the molecuk
Applying the relation between the number of molecules derived previouslj
there is fixed a significance for the weight of the molecule which for simple gasc
like hydrogen and oxygen is twice the atomic weight and for compound gasei
like methane and carbon dioxide, is equal to the atomic weight. Applyin
this to the Avagadro law, the weights of equal volumes of different gases mus
be proportional to their molecular weights, as equal volumes of all contain tl
same number of molecules.
Putting this in symbolic form and comparing any gas with hydrogen, as
its density, because it is the lightest gas of all and has well determined chara!
HEAT AND MATTER 449
teristics, requires the following symbols, denoting hydrogen values by the
subscript H.
Let m = molecular weight of a gas,
8 = density in Ibs. per cu.ft. = ,
then
and
<>
But as the molecular weight of hydrogen is for engineering purposes equal to 2
closely enough and hydrogen weighs .00562 Ib. per cu.ft. = S#, at 32 F and
29.92 ins. Hg,
Lbs. per cu.ft. = Si = .00281 mi ...... (648)
To permit of evaluation of Eq. (648) it is necessary that there be available
a table of molecular weights of gases and the atomic weights of elements from
which they are derived, and the values given at the end of this Chapter in Table
XLII are derived from the international table. As atomic weights are
purely relative they may be worked out on the basis of any one as unity, and
originally chemists used hydrogen as unity, but for good reasons that are of no
importance here, the custom has changed to Y& the value for oxygen as unity.
These atomic weights are not whole numbers but nearly so, therefore, for con
venience and sufficient accuracy the nearest whole number will be used in
this work and hydrogen be taken as unity except where experience shows it
to be undesirable.
The reciprocal expression to Eq. (648) can be set down, giving the specific
volume of a gas or its cubic feet per pound at 32 F. and 29.92 ins. Hg., as
follows :
Cu.ft. per ib.an ...... (649)
This is a most important and useful conclusion as applied to gases and vapors
[for which no better values are available, and in words it may be stated as follows:
The cubic feet per pound of any gas or vapor at 32 and 29.92 ins. Hg, is
equal to 355.87 divided by its molecular weight,
or
The molecular weight of any gas or vapor in pounds, will occupy a volume of
355.87 cu.ft. at 32 and 29.92 ins. Hg.
The approach to truth of these general laws is measured by the values
[given for specific volume and density at the end of this Chapter (a) experiment
ially derived and, (6) as derived from the hydrogen value by the law.
450 ENGINEEEING THERMODYNAMICS
Another and very useful relation of a similar nature is derivable from what
has been established, connecting the gas constant R with molecular weights.
ID
The general law PV = wRT when put in the density form by making ^ = TF
becomes
^ = RT. . (650)
Whence, comparing gases with each other and w^th hydrogen at the same
pressure and temperature
Pi
RH i *
and Si
which indicate that the densities of gases are inversely proportional to the
gas constants, or the density of any gas is equal to the density of hydrogen times
the gas constant for hydrogen divided by its own.
Inserting the values of density at 32 and 29.92 ins. Hg and of the gas con
stant for hydrogen, it follows that for any gas
Lbs.percu.ft^S^ ~ f . . ..... (653)
the reciprocal of which gives the specific volume at 32 F. and 29.92 ins. Hg, or 1
P
, (654)
Example. 1. Explanation and use of Chart, Fig. 134. This diagram is for the
purpose of finding the cubic feet, per pound, or pounds per cubic foot, of a gas at32 c
F. and a pressure of 29.92 ins. of Hg, if its volume or weight per cubic foot be known f
at any pressure and temperature. The curves depend upon the fact that the pounds
per cubic foot (8) vary directly as the pressure and inversely as the temperature.
That is
, T 29.92
The line of least slope is so drawn that for any temperature on the horizontal scale
its value when divided by 492 may be read on the vertical scale. The group of line,'
with the greater slope is so drawn that for any value on the vertical scale this quantit;
29 92
times ~ may be used on the horizontal scale. That is, the vertical scale gives th
HEAT AND MATTER 451
itio of densities as affected by temperature for constant pressure, while horizontal
3ale gives the ratio as affected by both temperature and pressure. A reciprocal
3ale is given in each case for volume calculations.
To find the pounds per cubic foot of gas at 32 F. and 29.92 ins. of mercury when
,s value is known for 90 and 13 Ibs. per sq.in. On the temperature scale, pass
ertically until the temperature line is reached, then horizontally until the curve
;>r 13 Ibs. absolute is reached. The value on the scale below is found to be 1.265,
) that the density under the standard conditions is 1.265 of the value under known
onditions. Had it been required to find the cubic feet per pound the process would be
precisely the same, the value being taken from the lower scale, which for the example
3ads .79, or, the cubic feet per pound under standard conditions is 79 per cent of
he value under conditions assumed.
Example 2. By means of the molecular weight find the density of nitrogen at
2 F. and 29.92 ins. Hg, and the cubic feet per pound for these conditions.
From Eq. (646)
81 mi 28 X. 00562
g^= , or 81 = .
lence 8 for nitrogen = .07868 pounds per cu.ft. and,
Prob. 1. Taking the density of air from the table, find the value of R for air, by
neans of Eq. (653) and compare its value with that found in Section 7.
Prob. 2. Compare the density of carbon monoxide when referred to 32 F. and
SO F. as the standard temperature, as found both ways.
Prob. 3. By means of their molecular weights find the density of oxygen, nitrogen
md carbon dioxide at 32 F. and 29.92 ins. Hg.
Prob. 4. What are the cubic feet per pound of acetylene, methane and ammonia
t 32 F. and 29.92 ins. Hg?
Prob. 5. An authority gives the following values for R. Compare the densities
ound by this means with the densities for the same substance found by the use of
he molecular weights.
Oxygen 48.1
Hydrogen 764.0
Carbon monoxide 55.0
Prob. 6. What will be the volume and density under standard conditions, of a
;as which contains 12 cu.ft. per pound at a temperature of 70 F. and a pressure of
o Ibs. per square inch absolute?
Prob. 7. What will be the difference in volume and density of a gas when con
jidered at 60 and 29.92 ins. of Hg, and at 32 F. and 29.92 ins. of Hg?
p 9. Pressure and Temperature Relations for Vapor of Liquids or Solids,
j'/aporization, Sublimation and Fusion Curves. Boiling and Freezingpoints
or Pure Liquids and Dilute Solutions. Saturated and Superheated Vapors.
e jfobstances may exist in one of three states, solid, liquid or gas, the latter being
generally called vapor when, at ordinary temperatures the common state is that
452 ENGINEERING THERMODYNAMICS
of liquid or solid, or when the substance examined is near the point of lique
faction or condensation, and just which state shall prevail at any time depends
on thermal conditions. Within the same space the substance may exist in two oi
these three states or even all three at the same time under certain special condi
tions. These conditions may be such as to gradually or rapidly make that parl
in one state, turn in to another state, or may be such as to maintain the relative
amounts of the substance in each state constant; conditions of the latter sort are
known as conditions of equilibrium. These are experimental conclusions, but
as in other cases they have been concentrated into general laws of which they
are but special cases. The study of the conditions of equilibrium, whether o
physical state or chemical constitution, is the principal function of physica
chemistry, in the pursuit of which the Gibbs phase rule is a controlling prin
ciple. According to this rule each possible state is called a phase, and th
number of variables that determine which phase shall prevail or how man>
phases may exist at the same time in equilibrium for one chemical substanc
like water, is given by the following relation, which is but one of the conclusion
of this general principle of equilibrium.
Number of undefined variables = 3 number of phases.
Now it is experimentally known that if water be introduced into a vacuun
chamber some of it will evaporate to vapor and that, therefore, water and it
vapor may coexist or the number of phases is two, but this does not state
or when equilibrium will be attained. The rule above, however, indicates tha
for this case there can be but one undefined or independent variable and, o
course, since the pressure rises more when the temperature is high than whei
low, the two variables are pressure and temperature, of which accordingly o
one is free or independent, so that fixing one fixes the other. In other word
when a vapor and its liquid are together the former will condense or the latte
evaporate until either pressure or temperature is fixed, and fixing tiie one th
other cannot change, so that the conditions of equilibrium are indicated
a curve to coordinates P and T, on one side of which is the vapor state anc
on the other that of liquid. Such a curve is the vapor pressuretemperature curv
of the substance, sometimes called its vapor tension curve, and much experi
mental information exists on this physical property of substances, all obtaine*
by direct measurement of simultaneous pressures and temperatures of a vapo
above its liquid, carefully controlled so that the pressure or the temperature i
at any time uniform throughout.
The conditions of equilibrium between vapor and liquid, defined by the vapo
tension curve extend for each substance over a considerable range of pressur
and temperature, but not indefinitely, nor is the range the same for each. A
the highpressure and temperature end a peculiar interruption takes place du
to the expansive effect of the temperature on the liquid and the compressiv
effect of the pressure on the vapor, the former making liquid less dense and tl
latter making vapor more dense, the two densities become equal at son:
pressure and temperature. The point at which this occurs is the " critical point
at which the equilibrium between liquid and vapor that previously existec
HEAT AND MATTER
453
nds and there is no longer any difference between vapor and liquid. This
k)int is a most important one in any discussion of the properties of matter,
,nd while difficult to exactly locate, has received much experimental attention,
nd some of the best values are given below in Table XXI for the pressure,
lensity, and temperature denning it, for the substances important in engineering'
TABLE XXI
THE CRITICAL POINT
Substance.
Symbol.
Critical Temp.
Critical Pres
sures.
Critical
Density
Water
at
4C=1.
Authority.
Criti
cal vol.
Cu.ft.
perLb.
Authority.
C.
F.
Atm.
Lbs.
per
Sq.in.
lydrogen
>xygen
Nitrogen
H 2
2
N 2
NH 3
NH;
C0 2
C0 2
H 2
H 2
H 2 O
H 2
H 2
H 2
243.5
118.1
146.1
+130.0
+131.0
+ 31.35
+ 30.921
+358.1
+364.3
+365.0
+374.
+374.6
+374.5
390.1
180.4
232.8
266.
267.8
88.43
87.67
676.4
687.7
689.
705.2
706.3
706.1
20
50i
35.i
115.
113.
72.9
77.i
194.61
200.5
294
735
515
1690
1660
1070
1130
2859
2944
3200
3200
.652
.442
.464
.452
.429
Olszewski
i Wroblewski
2 Dewar
1 Olszewski
2 Wroblewski
Dewar
Vincent and
Chappuis
Amagat
i Andrews
2 Cailletet and
Matbiaa
Nadejdini
Batteli
Cailletet and
Colardeau
Traube and
Teichner
Holborn and
Baumann
Marks
26.8
13.
Nadejdini
Batteli _.
Linmonia
iinmonia
/arbon dioxide. . .
Carbon dioxide. . .
Vater
Vater
Vater
Vater
Pater
Vater
To illustrate this discussion there is presented the vapor tension curves of
ater, ammonia and carbon dibxide to a large scale in chart form derived
rom the tabular data both at the end of this Chapter, while a small scale dia
ram for water is given in Fig. 135. These data are partly direct experimental
eterminations and partly corrections obtained by passing a smooth
urve representing an empiric equation of relation between pressure and tem
erature, through the major part of the more reliable experimental points.
se pressuretemperature points are very accurately located for ivater, the
rst good determinations having been made by Regnault in 1862 and the last
y Holborn and Baumann of the German Bureau of Standards in the last year,
lie data presented are those of Regnault corrected by various investigations
>y means of curve plotting, and empiric equations by Wiebe, Thiessen and
>chule, and those of various later observers, including Battelli, Holborn, Hen
ing, Baumann, Ramsay and Ypung, Cailletet and Colardeau, somes eparately,
but all together as unified by Marks and Davis in their most excellent steam
454
ENGINEERING THERMODYNAMICS
tables, and later by Marks alone for the highest temperatures 400 F. to the
critical point, which he accepts as being located at 706.1 F. and 3200 Ibs. square
Temperature in Degrees Fahr.
FIG. 135. Vapor of Water, Pressuretemperature Curve over Liquid (Water).
inch. In calculations the values of Marks and Davis, and Marks, will
accepted and used.
HEAT AND MATTER
455
Carbon dioxide and ammonia are by no means as well known as steam,
and the original data plotted, while representing the best values obtainable, must
be accepted with some uncertainty. A smooth curve Figs. (139) and (140)
has been drawn for each through the points at locations that seem most fair,
for both these substances and the values obtained from it are to be used in
alculations; these curves have been located by the same method as used by
^Earks in his recent paper and described herein later. The equalized values
,re given in the separate table at the end of the Chapter with others for latent
.1
c
HH
0<
GO
*H
G>
&j
02
3.05
.s
02
o>
i
i
<
/
/
/
Ice
/
/
/
A
ipor
/
S*
/

***?,
_
35 10 +15 +40
Temperature in Degrees Fahr.
FIG. 136. Vapor of Water, Pressuretemperature Curve over Solid (Ice).
heats and volumes, but while consistent each with the other are probably but
Ilittle more correct than values reported by others which are inconsistent.
The curves. and the equivalent tabular data are most useful in practical
(work, as they indicate the temperature at which the vapor exists for a given
i pressure, either as formed during evaporation or as disappearing during con
idensation, or the other way round, they indicate the pressure which must be
maintained to evaporate or condense at a given temperature.
Just as the vaporliquid curves indicate the conditions of equilibrium between
456
ENGINEERING THERMODYNAMICS
vapor and its liquid, dividing the two states and fixing the transition pressure
or temperature from one to the other, so also does a similar situation exist with
respect to the vaporsolid relations. In this case the curve is that of " sub
limation " and indicates the pressure that will be developed above the solid
by direct vaporization at a given temperature in a closed chamber. In Fig.
136 is plotted a curve of sublimation of vaporice, based on Juhlin's data,
Table XXII, which indicates that the line divides the states of ice from that
of vapor so that at a constant pressure, decrease of temperature will cause
vapor to pass directly to ice and at constant temperature a lowering of pres
sure will cause ice to pass directly to vapor.
TABLE XXII
JUHLIN'S DATA ON VAPOR PRESSURE OF ICE
Temperature.
Pressure.
C.
F.
Min. Hg.
Lbs. sq.in.
50
58.
.050
.001
40
40.
.121
.0023
30
22.
.312
.006
20
 4.
.806
.0156
15
5.
1.279
.0247
10
14.
1.999
.0386
 8
17.6
2.379
.0459
 6
21.2
2.821
.0544
 4
24.8
3.334
.0643
 2
28.4
3.925
.0758

32.
4.602
.0888
Likewise the liquid, water, may pass to the solid, ice, by lowering temperature
at a fixed pressure as indicated in Fig. 137, plotted from data by Tamman,
Table XXIII, and which becomes then the curve of "fusion."
TABLE XXIII
TAMMAN'S DATA ON FUSION PRESSURE AND
TEMPERATURE OF WATERICE
Temperature.
Pressure.
C.
F.
Kg. sq.cm.
Lbs. sq.in.
32.
1
1423
 2.5
27.5
336
4779.
 5.
23.
615
8747.4
 7.5
18.5
890
13658.8
10.0
14.
1155
16428.
12.5
9.5
1410
20055.
15.
5.
1625
23113.
17.5
.5
1835
26100.
20.
 4.
2042
27044.
22.1
 7.8
2200
31291.
HEAT AND MATTER
457
\
\
V
\
s
\
\
\
\
\
V
ater
\
\
\
Ice
\
\
I
\
\
\
\
Temperature in Degrees Fahrenheit
FIG. 137. Water Ice, Pressuretemperature Curve.
458
ENGINEERING THEEMODYNAMICS
These three curves plotted to the same scale meet at a point located at a pressure
of 4.6 mm. Hg. = .18 ins. Hg., and temperature +.0076 C. =32.01 F., ordi
narily taken at 32 F., which point is named the triple point, as indicated in
Fig. 138. The fact that the vapor pressure for water extends below freezing
point and parallels more or less that of ice indicates the condition of supercooled
02
!.l
Ice
Water
Tri
Vapor
)le Point
50
Temperature Degrees Fahr.
FIG. 138. Water Vapor Water Ice, Combined Curves of Pressuretemperature Rela
tion. The Triple Print.
water, one of unstable equilibrium instantly dispelled by the introduction of
a little ice at the proper stable state for this temperature.
Ordinary engineering work is not concerned with the entire range indicated
in Fig. 138 for any substance, but with the higher temperature ranges for some
and the low for others, with transition from solid to liquid state for metals
and similar solids and the transition from liquid to vapor for a great many, of
which water comes first in importance, then the refrigerating fluids, ammonia
HEAT AND MATTER 459
and carbon dioxide, and last certain fuels like alcohol and the petroleum oils
with their distillates and derivatives.
Meltingpoints, or the fusion temperature of such solids as are important,
are usually given for only one pressure, the standard atmosphere, as in ordinary
practice these substances are melted only at atmosphere pressure, and some
such values are given at the end of the Chapter in Table XLIII.
This is not the case, however, for boilingpoints, which must be denned
a little more closely before discussion. The vapor pressure curves indicate
that as the temperature of a liquid rises, the pressure rises also if the substance
is enclosed, but if the pressure were relieved by opening the chamber to a region
of lower pressure and kept constant, then the temperature would no longer
rise and boiling or ebullition would take place. The boilingpoint then is the
highest temperature to which the liquid and its vapor could rise under the
existing pressure. When not otherwise defined the term boilingpoint must
be taken to mean the temperature of ebullition for atmospheric pressure of
29.92 ins. Hg, and values for several substances are given at the end of this
Chapter in Table XLIV.
Vapor having the temperature required by the pressure of the pressure
temperature curve is known as saturated vapor, and this may be denned as
vapor having the lowest temperature at which it could exist as vapor, under
the given pressure. Vapors may, however, be superheated, that is, have
higher temperatures than saturated vapors at the same pressure, but cannot
so exist for long in the presence of liquid. Superheating of vapors, therefore,
implies isolation from the liquid, and the amount of superheat is the number of
degrees excess of temperature possessed by the vapor over the saturation
temperature for the pressure. In steam power plant work, especially with
turbines, it is now customary to use steam with from 75 F. to 150 F. of
superheat, and it might be noted that all socalled gases like oxygen and
nitrogen are but superheated vapors with a great amount of superheat.
It has already been mentioned that the saturated vapor pressuretemperature
curve of direct experiment is seldom accurate as found, but must be corrected
by empiric equations or smooth average curves, and many investigators have
sought algebraic expressions for them. These equations are quite useful also
in another way, since they permit of more exact evaluation of the rate of change
of pressure with temperature, which in the form of a differential coefficient
is found to be a factor in other physical constants. One of these formulas
for steam as adopted by Marks and Davis in the calculation of their tables
is given in Eq. (655), the form of which, was suggested by Thiessen:
0+459.6) lo gl ^ = 5.409(^212 )3.71XlOi[(6890 4 477*], . (655)
in which t = temperature F.; and p = pressure Ibs. sq.in.
This represents the truth to within a small fraction of one per cent up to 400
F., but having been found inaccurate above that point Professor Marks has
460 ENGINEERING THERMODYNAMICS
very recently developed a new one, based on Holborn and Baumann's
high temperature measurements, which fits the entire range, its agreement
with the new data being onetenth of 1 per cent, and with the old below 400
F., about onefifth of 1 per cent, maximum mean error. It appears to be the
best ever found and in developing it the methods of the physical chemists have
been followed, according to which a pressure is expressed as a fraction of the
critical pressure and a temperature a fraction of the critical temperature.
This gives a relation between reduced pressures and temperatures and makes use
of the principle of corresponding states according to which bodies having the same
reduced pressure and temperature, or existing at the same fraction of their
critical are said to be in equivalent states. The new 'Marks formula is given
in Eqs. (656) and (657), the former containing symbols for the critical
I pressure p c I and the latter g j vmg to them their numerical values,
( temperature T c abs. J
in pressure pounds per square inch, and temperature absolute F.
log 2i = 3.006854 (yl) [l + . 0505476^+. 629547 (^.7875 VI, . (656)
log p = 10.515354  4873.71 T *  .004050967 7 + .000001 392964 T 72 . . (657)
As the method used in arriving at this formula is so rational and scientific,
it has been adopted for a new determination, from old data to be sure, of the
relations between p and T for ammonia and carbon dioxide, so important as
substances in refrigeration, especially the former. According to this method
if PC and T c are the critical pressures and temperatures, both absolute, and
p and T those corresponding to any other point, then according to Van der
Waals,
/ m \
(658)
Accordingly, the logarithm of the critical divided by any other pressure, is
to be plotted against the quantity [(critical temperatur edivided by the tem
perature corresponding to the pressure) !], and the form of curve permits of
the determination of the function, after which the values of the critical
point are inserted. This has been done for NHs and C02 with the result for
NH 3
.... (659)
which on inserting the critical constants,
p c =114 atm. = 1675.8 Ibs. per square inch
T e = 727.4 F. absolute
which are the Vincent and Chap
puis values,
HEAT AND MATTER 461
becomes,
logp = 5.604221527.54r~ 1 171961T~ 2 .... (660)
For CO2 it was found that
(661)
which on inserting the critical constants,
p c = 77 atm. = 1131.9 Ibs. per square inch
which are Andrews' values,
T 7 ^ 547.27 F. abs.
becomes,
logp = 7.465814405.7657 7 ~ 1 + 1617501.3667 7 ~ 2 257086165.87067 7 ~ 3 . (662)
Curves showing the relation of reduced and actual temperatures and pressures
are given in Fig. 139 for ammonia and in Fig. 140 for carbon doixide.
For the past half century far more time and effort have been devoted to
making other formulas of relation of p to T for saturated vapor not only for
steam, but also for other vapors, than would have sufficed for accurate exper
imental determination, and as these help not at all they are omitted here. Equa
tions of physical relations can be no better than the data on which they are based,
and for the substances ammonia and carbonic acid the charts or formulas must
be used with a good deal of suspicion.
In all engineering calculations requiring one of these constants even for steam
no one is justified in using a formula; the nearest tabular or chart value must
be employed and it will be as accurate as the work requires. Time is at
least as important as accuracy, if not more so, for if too much time is required
to make a calculation in commercial work, it will not be made because of the
cost, indirect and approximate methods being substituted.
It is sometimes useful in checking the boilingpoint of some substance
little known, to employ a relation between boilingpoints of different substances
at the same pressure applied to a substance wellknown.
Let T a and TD be absolute temperatures of boiling for substances A and B under
same pressure;
Td and T b ' be absolute temperature of boiling for substances A and B under
some other pressure.
'Then,
^ = ^+c(7Yn) ........ . . (663)
lb 1*
Such equations as this are useful in finding the saturation curve of other sub
L stances from that for water, which is now so well established, when enough
Ipoints are known for the other substances to establish the constant c. Also
T '
the ratio ~ plotted against the temperature difference T b ' T b should give
462
ENGINEERING THERMODYNAMICS
^
Logarithm of (Critical Pressure f any other Pressure)
Vo ^ os bo o io 4^ as
x
X
x
^
/
s
X
//
/
X
X
X
?
X
y>
Jf
V
From Values of Wood o
" " " Dieterici
" Regnault +
" " Ledoux /
X s
s>
X
^
X
V
'
X
.2 .3 .4 .5
(Critical Temperature Divided by any other Temperature)!
,0
sure in Pounds Per SqJn.Abs^ ^
/
/
1
,
/
/
/
.,
?
/
y
x
CO ^V"
1
200
x
From Values of
Wood o
Dieterici
Regnault +
Ledoux /
y
X
^
Lc^
^r
.r 
^
40
10
200
60 110 160 210
Temperatures in Degrees Fahr.
FIG. 139. Ammonia Pressuretemperature Relations, for Saturated Vapor.
HEAT AND MATTER
463

OJ
/
1 5
/
/
Logarithm ol (.Critical Pressure^^ any other
'en
/
^
/
.
/<
/
s
/
m
x
x*
S
^
x
^
From Cailletet's Data
u Regnault's ^
" Stewart's Interpolation of Zeu
" Zeuner's Tabulation of Mollier
based on Amagat's Data o
^
tier's Data +
8 Formula
.
^
^
8
2
.1 .2 .3 .4 .5
(Critical Temperature Divided by any other Temperature)!
/
/
yUU
*
/
/
%
j
From Cailletet's Data
" Regnault's A
" Stewart's Interpolation of Zeuner's Data +
* Zeuner's Tabulation of Mollier's Formula
based on Amagat's Data o
/
&
,, Aft
/
m GOO
/
Pressure in Pounds
x
^
k
/
X
/
/?
X
/
j**
>^"
^
_
*
. r
100
50
Temperatures in Degrees Fahr.
464
ENGINEERING THERMODYNAMICS
a straight line, and if the line is not straight the experimental values may b
wrong or the law untrue. This procedure has been followed in Fig. 141, ii
checking the curves for CO2 and NH 3 against those for water, but it is impos
sible to say whether the discrepancies for C02 are due to a failure of the law o
bad experimental values, probably both, as the law holds poorly for water itself.
250
200
150
100
.6
Values of
T A '
"V
FIG. 141. Curves for CO 2 and NH 3 to Check the Linear Relation Eq. (663).
All of the preceding refers, of course, to pure substances, but in practi
work there are frequently encountered problems on solutions where lai
differences may exist compared to the pure liquids. Thus, for salts in water
is well known that addition of a salt lowers the freezingpoint, that more s
lowers it more, and it was first thought that the depression was in proporti
to the amount dissolved. This being found to be untrue, recourse was h
HEAT AND MATTER
465
again to molecular relations by Raoult, who announced the general law that
the molecular depression of the freezingpoint is a constant.
Molecular lowering of freezingpoint E' = = const.,
(664)
in which
A = depression of freezingpoint in degrees F. ;
w = weight dissolved in 100 parts of solvent;
m = molecular weight of substance dissolved.
From Eq. .(664) the freezingpoint for brines may be found as follows:
Freezingpoint of aqueous solutions = 32 (const.) X . .
(665)
As examples of the degree of constancy of the " constant " the following values
Table XXIV, taken from Smithsonian Tables are given:
TABLE XXIV
LOWERING OF FREEZING POINTS
g. Mol.
Molecular
Salt.
1000 g. H 2 O'
Lowering.
Authorities.
NaCl
.004
3.7
.01
3.67
.022
3.55
Jones
.049
3.51
Loomis
.108
3.48
Abegg
.232
3.42
Roozeboon
.429
3.37
.7
3.43
NH 4 C1
.01
3.6
.02
3.56
.035
.1
3.5
3.43
Loomis
.2
3.4
.4
3.39
CaCb
.01
5.1
.05
4.85
.1
4.79
.508
5.33
.946
5.3
Arrhenius
2.432
8.2
JonesGetman
3.469
11.5
JonesChambers
3.829
14.4
Loomis
.048
5.2
Roozeboon
.153
4.91
.331
5.15
1
.612
5.47
.788
6 34
466 ENGINEEKING THERMODYNAMICS
Just as the~pressure of dissolved substances in liquids lowers the freezing
point, so also does it lower the vapor pressure at a given temperature or raise
the boilingpoint at a given pressure. Investigation shows that a similar
formula expresses the general relation:
Molecular rise of boilingpoint = E = ^ = constant = 5.2, . (666)
when water is the solvent.
From Eq. (666) the rise of the boilingpoint is found to be
m
Rise of boilingpoint = 5.2 (667)
When liquids are mixed, such as is the case with all fuel oils and with
denatured alcohol, the situation is different than with salts in solution, and
these cases fall into two separate classes: (a) liquids infinitely miscible like
alcohol and water or like the various distillates of petroleum with each other,
and (6) those not miscible, like gasolene and water.
The vapor pressure for miscible liquid mixtures is a function of the pressure
of each separately and of the molecular per cent of one in the other when there
are two. This rule, which can be symbolized, is no use in engineering work,
because in those cases where such mixtures must be dealt with there will be
generally more than two liquids, the vapor pressure characteristic and molec
ular per cent of each, or at least some of which will be unknown.
When, however, the two liquids in contact or in fact any number are
nonmiscible they behave in a very simple manner with respect to each other,
in fact are quite independent in action. Each liquid will evaporate until its
own vapor pressure is established for the temperature, as if the other were not
there, and the vapor pressure for the mixture will be the sum of all the separate
ones. On the other hand the boilingpoint will be the temperature at which
all the vapor pressures together make up the pressure of say the atmosphere,
and this is necessarily lower than the highest and may be lower than the
lowest value for a single constituent. This action plays a part in vaporizers
and carburettors using alcohol and petroleum products. To permit of some
approximations, however, a few vapor tension curves for hydrocarbons and alco
hols are given later in the Section on vaporgas mixtures, and data on the vapor
pressure and temperature relations of ammoniawater solution are given in the
section on the solution of gases in liquids.
Example 1. Through how many degrees has ammonia vapor at a pressure of
50 Ibs. per square inch absolute been superheated, when it is at the temperature at
which steam is formed under a pressure of 100 Ibs. per square inch absolute?
From the curve of pressure and temperature of steam the temperature is 328 F.
for the pressure of 100 Ibs. From the similar curve for ammonia vaporization occurs
under a pressure of 50 Ibs. at a temperature of 22 F. Hence, superheat = 328
22=306F.
HEAT AND MATTER 467
Prob. 1. Three tanks contain the following liquids together: water, ammonia,
find carbon dioxide respectively, and at a temperature of 30 F. What pressure
iBxists in each tank? If the temperature rises to 70 F. how much will the pressure
;:ise in each?
Prob. 2. The pressure exerted by water vapor in the atmosphere when saturated,
that due to the temperature and is independent of the pressure of the air. The
itotal pressure read by a barometer is the sum of the air pressure and the water vapor
! pressure. What is the pressure due to each under a saturated condition for tem
peratures of 50 F., 100 F., 150 F., and 200 F., the barometer in each case being
29.92 inches of Hg?
Prob. 3. In order to secure a sufficiently high rate of heat transfer the steam in
fa, radiator must be at a much higher temperature than the room to be warmed. If it is to
be 150 above room temperature what must be its pressure for room temperatures
f 50 F., 60 F., 70 F., 80 F., and 125 F.?
Prob. 4. In one type of ice machine ammonia gas is condensed at a high pressure
d evaporated at a low pressure. What is the least pressure at which gas may be
ndensed with cooling water of 70 F., and what is the highest pressure which may
be carried in the evaporating coils to maintain a temperature in them of F.?
Prob. 5. Should carbon dioxide be substituted in the above machine what pressures
would there be in the condensing coils, and in the evaporating coils?
Prob. 6. How many degrees of superheat have the vapors of water, ammonia and
barbon dioxide at a pressure of 15 atmospheres and a temperature of 500 F.?
Prob. 7. Change the following pressures in pounds per square inch absolute \o
reduced pressures for water, ammonia, and carbon dioxide, 15 Ibs., 50 Ibs., 100 Ibs.,
500 Ibs.
Prob. 8. At the temperature of melting ice what will be the vapor pressure of ammonia
IJnd carbon dioxide? At the temperature of melting tin what will be the pressure of
{water vapor? At this same temperature how many degrees of superheat would
ammonia vapor under 100 Ibs. pressure have, and how many degrees superheat would
carbon dioxide vapor have under 1000 Ibs. pressure?
Prob. 9. If 10 Ibs. of common salt, NaCl, be dissolved in 100 Ibs. of water, what
will be the boiling point of the solution at atmospheric pressure, what the freezingpoint?
10. Change of State with Amount of Heat at Constant Temperature. Latent
Heats of Fusion and Vaporization. Total Heats of Vapors. Relation of Spe
cific Volume of Liquid and of Vapor to the Latent Heat. As previously explained,
i liquid boils or is converted into a vapor at constant temperature when the
pressure on the surface is constant. Then during the change of state the amount of
Iheat added is indirect proportion to the amount of vapor formed. The amount of
vapor to convert a pound of liquid into vapor at any one steady tempera
ture, is the latent heat of vaporization some values for which are given at the
end of this chapter in Table XLV, and it must be understood that this
natent heat is also the amount given up by the condensation of a pound of vapor.
I Latent heat is not the same for different pressures or temperatures of vapori
sation but is intimately associated with the volume change in the transition
jifrom the liquid to the vapor state. That this should be so, is clear on purely
Itational grounds because there is necessarily external mechanical work done
468 ENGINEERING THERMODYNAMICS
in converting the liquid to the vapor, since this is accompanied by a change
of volume against the resisting pressure at which the conversion takes place.
Thus, if
Vv = specific volume of the vapor in cubic feet per pound;
V L = specific volume of the liquid in cubic feet per pound ;
P = pressure of vaporization Ibs. per sq.ft. absolute.
Then
J Mechanical external work done dur 1 _p/y 17 ) ft lb (66*
{ ing vaporization of 1 Ib.
Of course, at high temperatures the volume of a pound of liquid is great*
than at low because of its expansion with temperature rise, and under the co
responding higher pressures the volume of a pound of vapor is less, becaus<
of the compressional effect of the pressure, than at low pressures, so that a
pressures and temperatures rise the difference VvV L becomes less and dis
appears at the critical point where it is zero. The latent heat being thus asso
ciated with a factor that becomes less in the higher ranges of temperature
and pressure may be expected, likewise to become less unless some other facto
tends to increase. All the energy of vaporization making up the latent hea
may be said to be used up in (a) doing external work as above, or (6) overcom
ing attraction of the molecules for each other. As at the critical point theri
is no molecular change and no external work, the latent heat becomes zer<
at this point.
This relation between latent heat and volume change was formulated b;
Clausius and Clapeyron, but Eq. (669) is generally known as Clapeyron'
equation :
Let L = latent heat ;
J = mechanical equivalent of heat = 778, or better 777.52, in sucl
cases as this;
T = absolute temperature of vaporization;
(i dP , .
j~i = rate of increase of vapor pressure per degree change of corre
spending temperature.
Then
This formula is used to calculate latent heat from the specific volumes of vapo
and liquid and from the curvature of the saturation curve when they are known
but as these volumes are especially difficult to measure, direct experimenta
determination of the latent heat should be depended upon to get numerica
values wherever possible. The formula will then be useful for the invers
process of calculating specific volumes from latent heats or as a means' o
HEAT AND MATTER
469
checking experimental values of both, one against the other. It is, however,
iust as useful to calculate latent heats from the specific volumes, and   of
&T
the vapor curve, when the latent heats are less positively determined than
jhe volumes or densities.
Another simpler relation of a similar general character exists and is useful
in estimating latent heats approximately for some little known substances
ike, for example, the liquid fuels, and in the use of which accurate physical
iata are badly needed. Despretz announced that
s nearly constant for all substances, and this was simplified by Ramsay and
Crouton on the assumption, first, that the volume of the liquid is very small
it ordinary temperatures and may be neglected, in comparison with the volume
pf the vapor, and second, that the volume of the vapor is inversely proportional
p the molecular weight m and directly proportinal to absolute temperatures
to that (Trouton's law)
>r
m constant = C
T
= c
m
(670)
,he constant c is given the following values by Young:
CO 2 c = 21.3
NH 3 c = 23.6
Hydrocarbons c = 20.21
Water and alcohols c = 26
For such substances as water and steam, the properties of which must be
bccurately known, general laws like the above are of no value compared with,
lirect experimental determination except as checks on its results, and even
hese checks are less accurate than others that are known.
These experimental data are quite numerous for water, but as generally
aade include the heat of liquid water from some lower temperature to the
oilingpoint. The amount of heat necessary to warm a pound of liquid from
emperature 32 F. to some boilingpoint, and to there convert it entirely into
apor is designated as the total heat of the dry saturated vapor above the origi
al temperature. This is, of course, also equal to the heat given up by the con
ensation of a pound of dry saturated vapor at its temperature of existence and
f the subsequent cooling of the water to some base temperature taken univer
Mly now as 32 F. in engineering calculations.
470 ENGINEERING THERMODYNAMICS
From observations by Regnault and formulated by him in 1863 the present
knowledge of the total heat of water may be said to date. He gave the
expression, Eq. (671), in which the first term is the latent heat at 32 and
one atmospheric pressure:
Total heat per pound dry saturated steam = H = 1091. 7 +.305 (J32). (671)
This was long used as the basis of steam calculations, but is now to be discarded
in the light of more recent experimental data, the best of it based on indirect
measurements by Grindley, Griessmann, Peake, who observed the behavior of
steam issuing from an orifice, together with the results of Knobloch and Jacob
and Thomas on specific heats of superheated steam, and in addition on direct
measurements by Dieterici, Smith, Griffiths, Henning, Joly. All this work
has been recently reviewed and analyzed by Davis, who accepts 1150.3J
B.T.U. as the most probable value of the total heat under the standard atmos
phere and the following formula as representing total heats from 212 up tol
400 F.
# = 1150.3+.3745(*212) . 0005500 212) 2 . . . (675
The Davis curve containing all the important experimental points and th
accepted line, extended dotted from 212 to 32, is presented in Fig. 142.
From the total heats given by this formula the latent heat is obtainab]
by subtraction, according to the relation,
Latent heat (L)= total heat of vapor above 32 F. (J^) heat of
liquid from 32 F. to boiling point (h), (673
in which the heat of the liquid is computed from a mean curve between Dieterici'
and Regnault's values, having the equation ft = .9983 .0000288 032) f
.0002133(<32) 2 . This is the basis of the values for latent and total heats in th
Marks and Davis steam tables referred to, and accepted as the best obtain
able today. From these tables a pair of charts for latent heat and total heat
dry saturated steam are given at the end of this Chapter.
The specific volume and density of dry saturated steam, given in the chart
and table are calculated, as this seems to promise more exact results than direc
experiment, the method of calculation involving three steps:
(a) From the pressuretemperature equation the ratio of is found b.
differentiation as follows:
log p = 10.5153544873.7ir~ 1 .00405096T+. 000001392964 7 12 ,
whence
^f = (^P^ .00405096+ .000002785928 T\ p.
HEAT AND MATTER
471
(6) From the latent heats the difference between specific volume of vapor
jind liquid, (V V V L ) is calculated by substituting (a) in Clapeyron's equation.
(c) From the Landolt, Bornstein, Myerhoffer tables for density of water
bhe volume V L is taken, whence by addition the volume of the vapor V v is found
For ammonia and carbonic acid there are no data available on total heats
by either direct measure or by the orifice expansion properties, and very few
1083,5/s
D
1069;
1112,4;.
] 126^8 H,
1117.C
I/
^1123.2
1155
1IJ7.
i:jy.
1144.8
1170
/
llWsi
1159,5
1184;
oX^
X
1166.
1173,
32 50 fcS 86 104 122 110 158 176 194 212 230 248 266 284
320 338 356 374 392
Temperature in. Degrees K.
FIG. 142. Total Heat of Dry Saturated Steam above 32 F. (Davis).
leterminations of the latent heat itself, so that the process that has proved so
iatisfactory with steam cannot be directly followed with these substances.
Vccordingly, a process of adjustment has been used, working from both ends,
Beginning with the pressure temperature relations on the one hand and specific
volumes of liquid and vapor on the other, the latent heat is determined by
plapeyron's equation and where this does not agree with authentic values an
idjustment of both latent heat and specific volume is made.
472 ENGINEERING THERMODYNAMICS
This process is materially assisted by the socalled Cailletet and Mathij
law of mean diameter of the curves of density of liquid and vapor, which
given in Figs. 143, 144 and 145, for water, ammonia and carbon dioxide, on
which the points are marked to indicate the source of information.
On each of these curves the line BD is the line of mean density, its abscissa
being given by the following general equation,
(674)
Of course, this mean density line passes through the critical volume B. For
these three cases this Eq. (674) is found to have the form,
(675;
For water s = 28.7 .015( 300) .000015(*300) 2 . (a)
For ammonia s = 20 .022(^30). (b)
For carbonic acid.. s = 33.1 . 0219(^+20) .00016(Z20) 2 . (c)
A more exact equation for water has been determined by Marks and Davis
in their steam tables and is
s = 28.424.01650(*320).0000132(320) 2 . , . . (676)
From the smooth curve, which has the above equation, the volumes and densi
ties of liquid and vapor that are accepted have been derived, and are presented
in chart form on a large scale and in tabular form at the end of the Chapter,
the values for water being those of Marks and Davis.
From these volume differences and the relation the latent heats have
been calculated and the newly calculated points are compared with experimental
values in Fig. 146.
The total heats are obtained by adding to the latent heat the heat of
liquid above 32 from 50 F. up to the critical point for C02 and to
150 F. for NHs, which include the working range for refrigeration.
These liquid heats have already been determined in Section 5 in discussing
specific heats.
Charts and tables at the end of this Chapter give the final values of total heat,
heat of liquid, latent heat, specific volume and density of dry saturated
vapor based on largescale plottings, without equations beyond that for the
pressuretemperature relations for saturated vapor, and the results are be
lieved to be as reliable as it is possible to get them without more experimental
data.
The properties of drysaturated steam are given in Table XLVII, and'
charts, A, B, C, D, E, F; the properties of superheated steam, in Table XLVIII;
drysaturated ammonia vapor in Table XLIX, and Charts G, H, I, J, K, L;
and drysaturated carbon dioxide vapor in Table L, and Charts M, N,
O, P, Q, R.
HEAT AND MATTER
473
I
Solid curves from data of
Marks and Davis
Dotted extension to reach
new critical temperature
as found by Marks
400
Upper Scale =Lbs.per Cu.Ft.
.025 .02
Lower Scale =Cu.Ft.per Lb.
Temperature in Degrees Pabr.
i
^
\
V
\
\
\
\
\
\
\
\
s
\
s.
>^
X
\
^
>>
^
^
*^,
b *^
Volume of Vapor in Cu.FU
FIG. 143. Specific Volume and Density of Liquid and Dry Saturated Vapor of Water.
ENGINEERING THERMODYNAMICS
250
J
i
5
f
#50
50'
(
150
1
?
j.
1 50
50
(
FIG.
"S
1
^J
T
Gtitic
\lPoi
.^
s*
\
*,
^,
^
N
^
\
\
\
\
/
\
>
<fe
j
\
,s
^
*i
\
S
,
\
\
T?
s
J
1
o
^
fl
\
XL
^
/
_v
/
\
V
M
\
V
rj
_li
s
f
iVr
V
t
\
^
I
V
\,
\
"S
r 1
H
V
r
\
V
H 1
n
H
\
.
3iiteri6i
\
Drewes
\\
^Lange 
V
\\\
%
^n'dr^ib
>
\
y
\
\
\
\\
_L
D
) 5 10 15 20 25 30 ' 35 40
.2 .1 .075 .05 .04 .03 .025
Upper Scale= Lbs. per Cu. Ft. Lower Scale =: Cu. Ft. per Lb.
H
1
l
t
R
D
\
\
\
\
\
T\
\
^JX
\
\
^
^
^
^s
^
**^
^
***'**
^.
^^^
*.
^^.
~~
>.
,
" .
* ,
~
_

^
,
* _
*.
"
_
) 5 10 15 gO
Volume of Yapor.'in Gubio Feet per Ppuad^.
144. Specific Volume and Density of Liquid and Dry Saturated Vapor of Amrxioni;
HEAT AND MATTER
475
o From Data of Amagat
" " Cailletet & Mathais
 Behn
3 .1 .05
Upper Scale = Lbs.pen Cu.Ft.
Lower Scale =Cu.Ft.per Lb.
80
Q 40
pera
S
Critical Poll
Volume of Yapgr to Cu.Ft.per Poun<^ ,
FIG. 145.Specific Volume and Density of Liquid and Dry Saturated Vapor of Carbon
Dioxide.
476
ENGINEERING THERMODYNAMICS
<!> ,
^^
^o
%
<
\
^o
I
cj
"? H
. ft
S" H
3
^^
fe
o
x^^,
II
S S
S i
3 c
c3 o
\;
v $
Q c
if
^.4.
^S  I
'
x
X
$
rt ' I
^ .2
?
\
*
bc
^Q n
\
{
\
o
\
8
^
x
X
x^^
oLedoux
<j> = Zeuner
^=Regnault
0= Beaton's Exp.
A = Von Strombeck Exp
X
X
x
\
00
\
r
~
\
h
N
0
\
o
o
\4K,
J
.
V
5
> en
8
HEAT AND MATTER
477
The volumes of drysaturated steam determined from
ompared with their pressures show that there exists
elation of the form for steam,
p(Vv Fz,) 1 ' 0646 = constant = 497,
the
an
tables when
approximate
(677)
tvhen pressures are in pounds per square inch and volumes are in cubic feet.
?his curve plotted to PV coordinates is called the saturation curve for the vapor,
t is useful in approximate calculations of the work that would be done by steam
xpanding so that it remains dry and saturated or the work required to compress
rapor such as ammonia under the same conditions. But as the specific volume
f liquid is generally negligible it may be written as one of the general class
: = constant, (678)
or which s = 1.0646 and constant = 497.
?his curve supplies a means for computing the work for wet vapors (not too wet)
is well as. dry, provided only that they at no time become superheated or change
heir quality, by using for V some fraction of the true specific volume repre
enting the dryness. The very fact that a great volume of vapor may be
ormed from an insignificant volume of liquid makes the saturation curve a
jseful standard of comparison with actual expansion and compression lines for
wet vapors.
. 147. Comparison of Steam Expansion Line of an Indicator Card with the Satura
ion Line for both Dry Saturated Vapor and for Vapor Constantly Wet at the Initial Value.
478 ENGINEERING THERMODYNAMICS
ing for the rods, the displacement volumes of each end of the cylinder will be
5990 cu.ins., and since the clearance volume is 4 per cent, the steam volume
will be 239.6 cu.ins. From the lefthand card it will be seen that the cutoff
was at point C, 16.5 per cent of stroke, hence the volume at C is (.165X
5990) +239.6 = 1228 cu.ins. It will also be seen from the card, that the pres
sure at C was 73.5 Ibs. per square inch absolute. From the curves or the
tables at the end of the Chapter 1 cu.ft. of dry steam at this pressure weighs
1228
.1688 Ib. and hence the weight of steam in this end of cylinder was 7^7, X. 1668
1720
or .1185 Ib. at cutoff. From the card it will also be seen that at the end of
the exhaust stroke, denoted by the point D, the pressure was 30 Ibs., at which ;,
the weight of 1 cu.ft. of dry steam is .0728 Ib., hence the weight of steam in the
a
cylinder was =^X. 0728 = . 01010 Ib., and the amount admitted was .1185
1728
.0101 = . 1084 Ib.
In as much as the two ends of cylinder are identical and as the cards from
both ends are practically the same, it may be assumed that the same weight !
of steam was in each end, or that .1084X2 = .2168 Ib. are accounted for by the
card per revolution, or .2168X150X60 = 1950 Ibs. per hour. There is then
the difference to otherwise account for, of 2600 1950 = 650 Ibs. per hour,
which can only have been lost by condensation. 2600 Ibs. per hour is 2600
^(150X60X2) = . 1442 Ib. per stroke, which with the .0101 Ib. left from pre
vious stroke would make .1543 Ib. in the cylinder at cutoff, and if it
were all steam its volume would be 1581 cu.ins., denoted by point E l 
on diagram. The ratio of AC to AE gives the amount of actual
steam present in the cylinder at cutoff, to the amount of steam and
water. The saturation curves CF and EG are drawn through C anc
E from tabular values and represent in the case of CF the volume^
which would have been present in the cylinder at any point of stroke had the
steam and water originally present expanded in such a way as to keep tht
ratio or dryness constant, and in case of EG, volumes at any point of the strok<
if all the steam and water originally present had been in form of stean
and had remained so throughout the stroke. Just as the ratio of AC to Al
shows per cent of steam present at cutoff, so does the ratio of distances of an?
points Y and Z, from the volume axis denote the per cent of steam present a
that particular point of the stroke. By taking a series of points along th
expansion curve it is possible to tell whether evaporation or condensation i
occurring during expansion, In this case the ratio,
= = .795, and IT = .86.
AE XZ
Hence, it is evident that evaporation is occurring since the percentage of steai
is greater in the second case.
HEAT AND MATTER 479
For some classes of problems it is desirable that the external mechanical
work be separated from the latent heat, and for this reason latent heat is
^iven in three ways :
(a) External latent heat,
(6) Internal latent heat,
(c) Latent heat total.
The external latent heat in footpounds is the product of pressure and volume
Change, or expressing pressures in pounds per square inch,
1
144
External latent heat = p(7yFz,) (679)
i
this is sometimes reduced by neglecting V L as insignificantly small as it really
s for most problems which are limited to temperatures below 400 for saturated
apor, in which case,
144
External latent heat = jpV v (680)
J
In all cases
Internal latent heat = L (Ext. Lat. Ht.) (a)
144
= L
= L~~pV r (c)
(681)
Fusion and freezing are quite similar to vaporization and condensation
i that they are constant temperature processes with proportionality between the
mount of substance changing state and the amount of heat exchanged. They
re different in as much as little or no volume change occurs. As there is so little
sternal work done it may be expected that there is little change in their latent
eats with temperature and pressure, but as a matter of fact it makes very
ttle difference in most engineering work just how this may be, because prac
cally all freezing and melting takes place under atmospheric pressure. There
oes not appear to be any relation established between heats of fusion like
lose for vaporization that permit of estimates of value from other constants,
direct experimental data must be available and some such are given for a
substances at the end of this Chapter in Table XL VI. As a matter of fact
ich laws would be of little use, and this is probably reason enough for their
Dndiscovery.
Example 1. Pigs of iron having a total weight of 5 tons and a temperature of
X)0 F. are cooled by immersing them in open water at a temperature of 60 F. If
lehalf of the water is evaporated by boiling, how much must there have been originally?
i The iron must have been cooled to the final temperature of the water, which
just have been 212 F. Also the heat given up by the iron will be the
480 ENGINEERING THERMODYNAMICS
product of its weight, specific heat and temperature difference, or, considering the
mean specific heat to be .15,
10,000 X (2000 212) X. 15 =2,682,000 B.T.U.
The heat absorbed by the water in being heated, considering its specific heat as unity
will be its weight times its temperature change and, since onehalf evaporates, the
heat absorbed in evaporating it will be half its weight times the latent heat, or
TF[(212 60) + X970] = 637^ B.T.U.
These expressions for heat must be equal, hence
Example 2. A tank of pure water holding 1000 gallons is to be frozen by meai
of evaporating ammonia. The water is originally at a temperature of 60 F. and
ice is finally at a temperature of 20 F. The ammonia evaporates at a pressure^
of one atmosphere and the vapor leaves the coils in a saturated condition. Howf
many pounds of ammonia liquid will be needed, how many cubic feet of dry saturated!
vapor will be formed, and how much work will be done in forming the vapor?
The heat to be removed is the sum of that to cool the water, the latent healij
of fusion of ice, and that to cool the ice, or for this case
[(60 32) +144+.5(32 20)1x8333,
8333 being the weight of 1000 gallons of water. Hence the B.T.U. abstract
amount to 1,466,608.
Each pound of ammonia in evaporating at atmospheric pressure absorbs 594 B.T.U.'. 
as latent heat and, therefore, 2470 Ibs. are needed. At this pressure each pound
vapor occupies 17.5 cu.ft., hence there will be 43,200 cu.ft. of vapor. At this saml*s
pressure the volume of a pound of liquid is .024 cu.ft., so that the work done per pounj
in evaporating the ammonia is 37,000 ft.lbs. and the total work is 915 XlO 6 ft.ll
Prob. 1. How much ice would be melted at 32 F. with the heat necessary
boil away 5 Ibs. of water at atmospheric pressure, the water being initially at
temperature corresponding to the boilingpoint at this pressure?
Prob. 2. What is the work done during the vaporization of 1 Ib. of liquid anhydroi
ammonia at the pressure of the atmosphere?
Prob. 3. From the tables of properties of anhydrous ammonia check the value I
the constant in Trouton's law given as 23.6 by Young.
Prob. 4. As steam travels through a pipe some of it is condensed on account
the radiation of heat from the pipe. If 5 per cent of the steam condenses how mua
heat per hour will be given off by the pipe when 30,000 Ibs. of steam per hour at;
pressure of 150 Ibs. per square inch absolute is passing through it?
Prob. 6. Brine having the specific heat of .8 is cooled by the evaporation of ammoi
In coils. If the brine is lowered 5 F. by ammonia evaporating at a pressure of 1 1
HEAT AND MATTER 481
bs. per square inch gage, the vapor escaping at brine temperature, how many pounds
f brine could be cooled per pound of ammonia?
Prob. 6. Steam from an engine is condensed and the water cooled down to a
emperature of 80 F. in a condenser in which the vacuum is 28 ins. of Hg. How many
lounds of cooling water will be required per pound of steam if the steam be initially
per cent wet?
Prob. 7. A pound of water at a temperature of 60 F. is made into steam at 100
bs. per square inch gage pressure. How much heat will be required for this, and what
be the volumes at (a) original condition; (b) just before any steam is made; (c) after
Jl the water has been changed to steam?
Prob. 8. A sand mold weighs 1000 Ibs. and 100 Ibs. of melted cast iron are poured
nto it. Neglecting any radiation losses and assuming the iron to be practically at
:s freezing temperature how much of the iron will solidify before the mold becomes
1 the same temperature as the iron?
Prob. 9. How many pounds of ice could be melted by heat given up by freezing
.0 Ibs. of lead?
11. Gas and Vapor Mixtures. Partial and Total Gas and Vapor Pressures,
/blume, Weight, and Gas Constant Relations. Saturated Mixtures. Humidity.
)ne of the characteristic properties of gases distinguishing them from liquids,
ind which also extends to vapors with certain limitations is that of infinite
xpansion, according to which no matter how the containing envelope or volume
)f the expansive fluid may vary, the space will be filled with it at some pres
sure and the weight remain unchanged except when a vapor is brought to
;ondensation conditions, or the pressure lowered on the surface of a liquid
vhich will, of course, make more vapor. A given weight of gas or vapor (within
imits) will fill any volume at some pressure peculiar to itself, and two gases,
wo vapors, or a vapor and gas, existing together in a given volume, will fill it
it some new pressure which is the sum of the pressures each would exert sepa
ately at the same temperature (if nonmiscible) . This fact, sometimes des
gnated as Dalton's Law, permits of the derivation of equations for the rela
ion of any one pressure, partial or total, to any other total or partial, in
:erms of the weights of gas or vapor present, and the gas constants R. It
ilso leads to equations for the various constituent and total weights in
ierms of partial and total pressures and gas constants. Such equations sup
)ly a basis for the solution of problems in humidification and drying of air, in
tarburetion of air for gasolene and alcohol engines, or of water gas for illumina
ion, and are likewise useful as check relations in certain cases of gas mixtures
uch as the atmospheric mixture of nitrogen and oxygen, producer gas or gase
*us combustibles in general.
Let wi, w 2 and w x be the respective weights of the constituents of a mixture;
" w m = Sw be the weight of the mixture;
" PI, P2, Px be the respective partial pressures of the constituents;
" P m = SP be the pressure of the mixture;
" Ri, R 2 , Rx be the respective gas constants;
" R m be the gas constant for the mixture.
482 ENGINEERING THERMODYNAMICS
Then if wi Ibs. of one, and w 2 Ibs. of another gas or vapor at temperature
T m occupy the volume V m cubic feet together,
Vnfi^wiRiT*, ....... (a)}
and (682)
V m P2 = w 2 R 2 T m) ........ (b)\
whence
V m (Pi+P2) = (wiRi+W2R<i)T m , ...... (683)
or, in general,
2P=2(wR)T m ......... (684)
Or putting
2P = P m , . .. ......... (685)
and
or R m = ^^ 9 . . . (680)
then
P m V m = w m R m T m , . . . . . .... (687)
so that the mixture will behave thermally quite the same as any one gas with!
si^ch exceptions as may be due to a different gas constant R m .
Dividing Eq. (682a) by Eq. (683) or (684) gives
Pi = .wiRi _ wiRi I;
P m wiR l +w 2 R 2 2(wR)'
which gives the ratio of any partial pressure to that for the mixture in terms
of the individual weights and gas constants. Hence
W m R m >
which gives the ratio of any partial pressure to that for the mixture in terms Oj
its own weight and gas constant and those for the mixture.
It is possible to express the ratio of weights as a function of gas constants
alone which will permit of a third expression for the partial pressures in terms
of gas constants without involving any weights. For two gases
Whence
Wi W 2
HEAT AND MATTER 483
But from Eq. (686)
that
R 2
nd
is the ratio of partial to total weights in terms of gas constants. On sub
titution in Eq. (689),
(69)
which graves /ie rafa'o o/ partial pressures of two gases or vapors to that for the
I mixture in terms of the individual gas constants and that for the mixture, and
I a similar expression can be found for more than two gases. The ratio of any
i one partial, to the total weight can also be found from Eq. (689) in terms of its
i gas constant and partial pressure, and the mixture gas constant and pressure,
! from Eq. (691) in terms of the gas constants for the constituents and for
'the mixture. This ratio of partial to total mixture weight gives the fractional
composition by weight.
It is sometimes necessary to know the volume relations in a mixture of two
gases existing at the same pressure or two vapors or a vapor and gas, such, for
example, as air and water vapor. In this case two different volumes existing
together at a common temperature and pressure together make up a mixture
volume equal to their sum. Using similar symbols
(693)
where Vi and V 2 are the volumes occupied by the two constituents respectively
when at a mixture pressure P m and temperature T m , whence for the mixture
T m ..... (694)
or
(695)
484 ENGINEERING THERMODYNAMICS
These Eqs. (694) and (695) are identical in form with (683) and (684) except that
V replaces P, and V, P, so that all equations just derived also apply to volum<
as the volume proportion will be identical with pressure proportions. F<
convenience of reference these may be set down.
From Eq. (688),
g = ^, (696)
which gives the ratio of any partial volume, to that for the mixture in terms oj
the individual weights and gas constants.
From Eq. (689)
^=^^, (697)
r fM
which gives the ratio of any partial volume to that of the mixture in terms of its
own weight and gas constant and those for the mixture.
From Eq. (692)
Vl = Rl(R2~Rm)
V m R m (R2RiY
(698)
which gives the ratio of any partial volume to that of the mixture in terms of the
individual gas constants and that for the mixture.
The volumetric composition of air is given by Eq. (697) or its equal
numerically Eq. (692), and since the partial pressure of oxygen and nitrogen
in air are 78.69 per cent and 21.31 per cent, these are its volumetric per cents.
When one of the constituents is a vapor, all the preceding applies, provided
the condition of the vapor is such that at the temperatures assumed it is not
near the condition of condensation, but then the relations become more definite
since the partial pressure of the vapor is fixed by the temperature. In practical
work with gas and vapor mixtures the failure^of the perfect gas laws near the
condensation condition is ignored and they are assumed to be true for the very
good reason that there is no other way as good, to get numerical results.
All liquids, and many, if not all solids will, if placed in a vacuum chamber,
evaporate until the pressure has reached a certain value depending on the tem
perature, at which time the liquid and its vapor are in equilibrium, and evapo
ration may be said either to cease or proceed at a rate exactly equal to the rate at
which vapor condenses, or more precisely, at equilibrium the weight of vapor
in the vapor form remains constant. The weight of vapor that will rise over a liquid
in a given space depends on the temperature and pressure which are related
HEAT AND MATTER
485
Temperature,Deg. Cent.
80 *P , 5? , 6 70 80 90 100 110 . l2jfl
I ' H rl r 1 "l t^t 'I ' 1 1 I I ' r 1 r 1 ! '. ' .
800
750
50
30 40 60 80 100 120 140 160 180 200 220 240
Temperature,Deg. Fahr.
148 Vapor Pressure of Hydrocarbons and Light Petroleum Distillates of the Gasolene
Class.
486
ENGINEERING THERMODYNAMICS
in the socalled vapor tension or vapor pressure tables and curves, such as,
shown in Figs. 148, 149 and 150, for some liquid fuels or as given in the pre<;
vious section for water. At any fixed temperature the vapor will coni
tinue to rise until it exerts jts own vapor pressure for the temperature
the process being often described as evaporation without ebullition. If th(
liquid or solid be introduced into a chamber containing dry gas the evaporaj
tion will proceed precisely the same as in the vacuum until the pressure ha$!
vo
Temperature Deg. Centigrade
30 40 50 60 70
00
i.O
100
ou
80
100 120 HO
Temperature Deg. Fahrenheit
100
180
200
FIG. 149. Vapor Pressure of Heavy Petroleum Distillates of the Kerosene Class.
risen by an amount corresponding to the vapor pressure for the temperature
because each substance exerts the pressure it would if alone occupying th
volume; when they both occupy the same volume the pressure will be thei
sum and equal to the pressure of the gas alone, added to the vapor pressure fo
the same temperature. There is one important practical condition, and tha
is, time enough for the completion of the process of evaporation which proceed
very slowly toward the end. F time .enough is allowed the vapor pressure
establish itself and the gas is said to be saturated, and this is an importan
HEAT AND MATTER
487
special case of gasvapor mixtures. It is the condition in which the gas may
be said to carry the maximum weight of vapor possible for the total pressure
Temperature, Deg. Centigrade
30 40 50
J 1 1 1 I L
70 ' 80 ' 90 ' 100 ' 10 ' 120 130 140 150 160 170
Temperature Degrees, Fahrenheit
FIG. 150. Vapor Pressure of the Alcohols.
and temperature. The gas in contact with the liquid may carry less vapor
if it has not been in contact long enough aVthe given temperature, and a gas
488 ENGINEERING THERMODYNAMICS
no longer in contact with the liquid may carry less, because, (a) of insufficient
time of original contact; (6) of condensation of some it originally carried; (c)
of a rise of temperature after leaving the liquid. To all such general cases the
equations above apply without change, but for the special case of saturated
mixtures they have a simpler form.
Let P v be the vapor pressure of the liquid for temperature T, which is the
partial pressure of the vapor in a saturated vapor gas mixture ;
P g be the partial pressure of the gas at same temperature.
Then for a gas saturated with vapor at temperature T, Eq. (689),
/Weight vaporX _ /Vapor pressureX iR for mixtureX
\Weight mixt./ " \Mixture press!/ X \ R for vapor / ' ' '
But according to Eq. (646) and Eq. (651),
'R for vaporX _ /Density of mixtureX _ /Mol. wt. of mixt.X (
ft for mixt./ ~ \ Density of vapor / " \Mol. wt. of vapor/ ' '
whence
Wy _P V m v
w m P m m m ' .....
Also
C Plfni' ' ' ' ' ; ' ' ' (7(
and
w v
The presence of water vapor in the atmosphere, and problems connected with
it, constitute a specific case of vaporgas mixture, subject to the foregoing laws.
This subject has been given most attention by the United States Weather
Bureau; tables have been prepared for ready computation and for certain
problems for which only experimental data or empiric formulas afford solution.
Air is said to be " saturated with moisture " when it contains the saturated
vapor of water. It might be better to say that the space is saturated since the
presence or absence of the air has no effect upon the water, vapor other than im
posing its temperature or imparting heat to the water vapor, and also that the
air retards the diffusion of water particles. The weight of saturated aqueous
vapor per cubic foot depends only on the temperature, and not on the presence
of air.
If the space contains only a certain fraction of the weight of aqueous vapor
corresponding to saturation, that fraction is called the " relative humidity "
or degree of saturation, and the corresponding percentage, the per cent of
saturation. If air containing saturated water vapor be cooled ever so little
HEAT AND MATTER 489
some of the vapor will be condensed and appear in the liquid form. If air is
cooled at constant pressure, from a given initial condition, the degree
of saturation approaches unity, and finally reaches that value at a temperature
called the " dew point " corresponding to the initial condition. At this tem
perature the condition of saturation has been reached and any further cooling
will cause the precipitation of liquid water, as occurs in the formation of dew,
rainclouds or fog.
A space or body of air carrying water vapor in smaller quantity than that
;of saturation, in reality contains superheated steam. If the vapor density
iand the temperature of the mixture be known, the degree of superheat may be
[ascertained from the temperature of the mixture, and the temperature corre
I spending to saturated water vapor having a pressure equal to the partial pres
jsure of the vapor in the mixture.
Humidity of atmospheric air is ordinarily determined by an instrument
[called the psychrometer, which consists merely of two thermometers, one with
 a bulb exposed directly to the air and the other covered with a piece of wick
[which is kept moist with water. The two are mounted together so that they
[can be whirled or swung about in the air until a stable condition has been
[reached. The drybulb thermometer should record the temperature of the air.
[The wetbulb thermometer will record something lower* than the air tempera
jture, dependent upon the rate at which evaporation takes place, since the process
jof evaporation abstracts heat. Were there no other influence, this process of
[evaporation would continue till the temperature of the wet bulb became that
I of the dew point. The temperature of the wetbulb thermometer never falls
[to the dew point, however, because of conduction of heat between the cold
bulb and the warmer surrounding air. From extensive experiments conducted
I by the U. S. Weather Bureau, Professor Ferrel has devised the following formula
jfor the vapor pressure, h in ins. of mercury corresponding to given readings
{of the wet and drybulb thermometers, U and t w degrees F.
(704)
where h b is barometric height in inches, after applying all corrections, and h' is
pressure of saturated water vapor, in inches of mercury, corresponding to the
temperature t w .
The relations shown by this formula are expressed graphically in much
more convenient form in the curves of Figs. 151 and 152, devised by
Prof. H. L. Parr. The use of the curves is best illustrated by an example:
f the drybulb reading is 75 F. and the wetbulb 65 F., find the dew point.
The difference of wet and drybulb temperatures is 10. From 10 at the top of
;he diagram (B) Fig. 151 project downward, and from 75 air temperature at the
.eft of diagram project to the right to the intersection, where the dew point is
Iread by interpolation between the contour curves at (C) to be 59.5 F. These
?urves are drawn for a barometric pressure of 29.92 ins. (standard) and will
490
ENGINEERING THERMODYNAMICS
IOC
Difference in Temperature: Wet and Dry Bulbs: Degrees Fahrenheit
2 4 6 8 10 12 14 1 1
2 34 5 6 7 89 11
Difference in Temperature: Wet and Dry Bulbs: Degrees Centigrade
FIG. 151. Relation between Wet and Dry Bulb Psychrometer Readings and Dew Point for
Air and Water Vapor.
HEAT AND MATTER
491
\ \ \ \ l\ l\ IX IX
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
\
X
&
X
X
X
X
100
90
80
70
60
be
&
:> OJ
I
I
a
40
30 :
10 20 30 40 50 60 70.
Degree of Humidity, Per Cent
FIG. 152. Relation between Humidity and Weight of Moisture per Cubic Foot of
Saturated Air.
492 ENGINEERING THERMODYNAMICS
not apply correctly, when the barometer is not equal to this, though with fair
approximation, so long as the difference in barometer is not great. Where there
is much departure the original the formula must be used. Fig. 152 gives
weight of aqueous vapor per cubic foot of mixture, in grains (7^5 lb.) and
also the degree of humidity. The temperature of the dew point 59.6 F, is
located at (C") on the righthand side of Fig. 152. Interpolation between the
ends of the contours for weight, gives 5.6 grains per cubic foot. On the same
scale the temperature of the air, F., is represented at point (A) 75, project
ing to the intersecting point D and down to the bottom of the diagram gives
on the scale for degree of humidity, 60 per cent.
Example 1. By means of the relation of gas constants find the proportion of nitrog
and oxygen in the air.
R for nitrogen is 54.92 and for oxygen 48.25 and for air 53.35. From Eq. (698)
V m R m (RNRoY
which, on substituting the above values for R N , RQ and Rm gives
j^ 48.25 (54.9253.35)
V m "53.35 (54.9248.25) '
or air is 21.3 per cent oxygen by volume.
Example 2. At what temperature will air containing ^ lb. of water per pound of l
dry air at atmospheric pressure be saturated?
If the vapor pressure be known, the temperature may be found from tables, i
From Eq. (701)
Wg
or
For air m g = 28.88 and for water m p = 18,
hence substituting those values
, and P c +P,=760mm. of Hg.
(760P c )(28.88x.5)
18x1
or
which corresponds to a temperature of 172 F.
HEAT AND MATTER 493
Example 3. A pound of alcohol requires 9.06 Ibs. of air for a proper combustible
mixture for gas engines. At what temperature will these proportions contitute a
saturated mixture?
From Eq. (701)
^
~
For alcohol m 46, for air m g = 28.88, and P V +P =7W mm. of Hg for atmospheric
pressure. Substituting these values in the above equation
(760P,)28.88
46X9.06
From the curve of vapor tension of alcohol, the temperature corresponding to 49 mm.
of Hg is about 72 F. ,
Prob. 1. Air at 80 per cent lumidity, atmospheric pressure and 70 F, is cooled
to 40 F. How much water will be thrown down per 1000 cu.ft. of moist air.
Prob. 2. The same air is compressed adiabatically to five atmospheres, and again
cooled to 40 F. at this pressure. How much moisture per 1000 cu.ft. of moist air
fwill be separated out when the tenperature becomes 70 F, and bee much at
J40 F.
Prob. 3. What will be the weight of water in a pound of air and water vapor if
[the value for R for the mixture is taken as 55.25, for air as 53.35 and for water vapor
las 91?
Prob. 4. At what temperature will air containing its own weight of water vapor
be saturated at atmospheric pressure?
Prob. 5. An internal combustion engine uses a saturated mixture of air and gasolene
jvapor in which ratio of air to gasolene is 15.3. Considering the gasolene to be hex
lane, at what temperature will the mixture be?
Prob. 6. Should kerosene regarded as decane, C 10 H 2 2, be substituted for gaso
e, in the above problem what would be the change in temperature of mixture,
uming it still to be saturated?
Prob. 7. Air containing moisture equal to one per cent of the weight of the air
ijilone is at a temperature of 150 F. How much is the water vapor superheated?
iVhat is the humidity?
Prob. 8. The reading of a dry bulb of a psychrometer is 90 F. and of the wet
)ulb 70 F. By means of curves of Figs. 151 and 152, find the dew point, relative humid
ty, and grains of water per cubic foot of air.
12. Absorption of Gases by Liquids and Adsorption or Occlusion by Solids.
Relative Volumes and Weights with Pressure and Temperature. Heats of
Absorption and of Dilution. Properties of Aqua Ammonia. In the attainment
)f high vacuua in steam condensers, separate removal of considerable quantities
df noncondensible gases is found necessary by means of dry vacuum pumps,
i fact that proves in a practical way the freedom with which the boiler water
lad absorbed gases. These gases for otherwise pure water are carbon dioxide
iind air, probably mainly air, but may include many others, notably the
494 ENGINEERING THERMODYNAMICS
products of organic decomposition especially when condensing water is taken
from the neighborhood of sewers, as is generally the case when power plants
are located on city water fronts. To a very much greater extent, however,
is ammonia soluble in water, the latter being capable of taking up about 1000
volumes of ammonia at and one atmosphere, against about 30 volumes of air,
and onefiftieth of a volume of hydrogen. It is the freedom of solution of am
monia in water that makes the process useful as a means of removing
anhydrous ammonia from the cooling coils in mechanical refrigerating plants,
as a substitute for the mechanical removal by piston compressors.
In all cases the solubility of gases in liquids decreases with rise of tem
perature, a fact associated with the separation of gases from boiler feed
water during their heating in feedwater heaters, economizers and the boiler
itself. This property is also depended upon to free the aqua ammonia that
has absorbed its ammonia charge from the cooling coils, of the amount so
taken up, by heating the solution in a separate chamber from which the am
monia vapor escapes to the ammonia condenser to become liquid anhydrous
while the weak liquor returns to the absorber for a new charge. To permit of
the calculation of the quantity of liquor to be circulated, in order that a given
amount of anhydrous ammonia may be absorbed from the cooling coils and
delivered later by heating to high temperature to a condenser, requires accu
rate data on the maximum possible ammonia content of solutions at various
temperatures and pressures. Rise of temperature always will reduce the
gas content of the solution if originally saturated, but the volume dissolved
is independent of pressure for slightly soluble gases like nitrogen or hydro
gen, the weight dissolved, of course, being greater and directly proportional to
pressure at a given temperature by reason of the increased density.
This law of independence of volume and pressure or proportionality
of weight to pressure, is known as Henry's law, and is hardly true for gases
as freely soluble as ammonia, probably due to some action between water and
the gas, equivalent to that studied by Thomsen for solids, which tend to form
hydrates of various kinds. For such gases as are soluble by weight in propor
tion to pressure, it is not the total pressure of the solution that is significant,
but the partial pressure of the gas alone, so that the amount of mixed gases
like air dissolved in water will depend, for the oxygen part, on the specific
solubility of oxygen and its partial pressure in the air, which is approximately
onefifth that of the air, and the same is true for the nitrogen. Thus, in examin
ing the solubility conditions for ammonia in water, while in practice the total
pressure only is known, it is to the separate pressure of the ammonia that the
amount dissolved must be referred in any attempt to establish a law of
relation.
Just as gases dissolve in liquids, so are they absorbed by solids, though in this
case the process is described as one of adsorption or occlusion. This phenomenon
is now being studied in connection with coal, which it is found more or less
freely absorbs air, the oxygen of which under comparatively low temperatures
unites with the coal causing spontaneous combustion if the heat is conserved
HEAT AND MATTER
as in
Most
a pile or its liberation in any way accelerated by heating or otherwise,
investigations of adsorption or occlusion have been made with charcoals,
32 50
70
110 130 150 170 190
Temperature in Degress Fahrenheit
210 230
TIG. 153. Ammoniawater Solutions, Relation between Total Pressure and Temperature
(Dotted Lines Mollier Data).
,he more dense varieties of which have greater adsorptive power than others.
Hie quantity of different gases adsorbed is believed by Dewar to be the same
n volume per unit of charcoal when each is held at its own condensation tern
496
ENGINEERING THERMODYNAMICS
perature. The quantity increases with rise of pressure but not in proportion,
and decreases rapidly with rise of temperature and a curve showing the tem
130
120
10 ll. 15 20 23.3925 30
Per Cent by Weight of Ammonia in Solution.
FIG. 154. Ammoniawater Solutions, Relation between Total Pressure and Per Cent NH
in Solution.
perature and pressure at which equal volumes are adsorbed is similar in form
to a vapor tension curve.
In the establishment of the properties of aqueous solutions of ammonii
HEAT AND MATTER
497
progress was for many years very slow and the early experimental data of
Abegg and Rosenfeld, 1903, for weak solutions and low pressures, that of Roscoe
and Dettman, 1859, for highly concentrated solutions under low pressures at
350
325
300
275
250
S 225

200
150
I
 US
100
75
50
25
^X
Bill
uareIn
5 10 11.8 15 20 23.3925 30 '& &
Percent by Weight of Ammonia in Solution
FIG. 155. Ammoniawater Solutions, Relation between Temperature and Per Cent NH 8
in Solution.
32 F., together with that of Sims, 1862, for four temperatures from 32 F.
to 212 F., failing to check and having many gaps, has been more recently supple
mented by Perman, 1903, and Mollier, 1908. With these new data it has been
Possible to graphicajly fill in the data of the unexplored region with reasonable,
498 ENGINEERING THERMODYNAMICS
though not yet satisfactory accuracy. The results are given in three diagrams,
grouping the three variables of pressure, temperature and concentration in
pairs, Fig. 153, giving pressuretemperature, Fig. 154, pressureconcentration
and Fig. 155, temperatureconcentration, from which can be read off with reason
able accuracy any quantity needed in calculations and from which Table
LI, has been prepared. In this table the lower numbers are new, and the
upper those as given by Starr several years ago and since used by engineers
engaged in refrigeration, as standards.
These data refer to he equilibrium conditions of the solution, and in using
them for practical problems care must be taken to avoid applying them to other
conditions, for example to solutions that are not homogeneous, or in which
there has not been sufficient time for the establishment of equilibrium. Over
the surface of such solutions there exists a mixture of water vapor and
ammonia, each exerting its own partial pressure and the sum of the partial
pressures making up the pressure of the solution; it must not be assumed,
however, that the partial pressures are those of the pure substances, as this is
a true solution and not a simple mixture. Moreover, there is no certainty that
it is always a solution of just ammonia in water, as there is a reasonably good
possibility that hydrates of ammonia may form, which would further complicate
the relative pressures of the two constituents.
It is from Perman that the most accurate data on the composition of the
equilibrium vapor mixture have come, and he calculated the partial pressures
from the composition of the mixture determined by analysis and using Dalton's
law,
Partial pressure of ammonia _ Volume of ammonia
Total pressure "Total vapor volume'
No success has yet been met with, in attempting to express a general relation
between partial pressures and the two variables, pressure and temperatures,
so Perman's values are given as found in Table LII, and for the unex
plored region it is not possible to do better than make a guess at a needed value.
As a check on the Perman values and to assist in estimating needed values for
other ranges the sum of the two partial pressures is given and compared with
the accepted values for the total pressure, and it should be noted that as the
partial pressure relations give the volumetric composition of the vapor mix
ture, it is unfortunate that the data are not extended to at least 35 per cent
concentration to cover the solutions in the generator of the absorption refriger
ating machine, from which both ammonia and watervapor are discharged
in as yet unknown relative amounts and which must be separated as completely
as possible before condensation.
Any change in the ammonia content of a solution is thermal in character
and is, therefore, accompanied by heat changes. When water absorbs ammonia
heat is liberated, as is also the case when ammonia in solution is diluted with
more water, the latter being really a further absorption in the fresh water of
the ammonia already contained in its solution being diluted by it. Likewise,
HEAT AND MATTER 499
when ammonia is absorbed by an ammonia solution heat is also liberated,
but heat is absorbed by a solution from which ammonia is escaping, as in
evaporation of liquids. There are three cases of the exothermic process, each
with an endothermic inverse and these are:
(a) Absorption of ammonia by pure water.
(6) Absorption of ammonia by ammonia solution.
(c) Dilution by water of an ammonia solution.
Data on the amounts of heat liberated in these cases are not sufficient
to establish firmly any general law of change, but are sufficient to give an
approximation. The first important fact in this connection is that the heat
liberated per pound of ammonia when pure water absorbs ammonia depends on
the amount of water. One pound of ammonia absorbed in a little water gives
out a little heat, more is liberated when more water is present, but when the
amount of water is large, put at 200 times the weight of ammonia by Thomsen,
the heat of absorption is constant and does not increase beyond this point.
It may easily be, however, that the point is reached with fifty water .weights, or
that some heat continues to be generated for any amount of water to infinity,
but so small in quantity as to be impossible to measure in the great weight
of liquid present. For example, if one B.T.U. were liberated in 100 Ibs.
of solution, the temperature rise would be somewhere near to Vioo F., and to
detect this in the presence of radiation and conduction influences and make
allowance for the heat of stirring would be difficult.
For engineering purposes, however, it may be accepted that the heat of absorp
tion of a pound of ammonia is constant if the weight of water is large, and its value
was fixed at 926 B.T.U. per Ib. by Favre and Silberman and later redetermined
by Thomsen at 8430 calories per gram molecule of NHa, which is equivalent to
9 8430\
X ) = 893 B.T.U. per pound. This value is accepted and defined as the
5 17 /
heat of complete absorption for the want of a better term, and in view of the
desirability of distinguishing it from the heat of absorption in small amounts
of water or in solutions already containing appreciable amounts of ammonia,
which will be designated as heat of partial absorption.
Experiments have further established another important relation between
the heats of dilution of solutions and their original strength. According to
this, solutions behave like ammonia itself with respect to pure water and lib
erate a little heat when a little water is added, more with larger amounts,
attaining a constancy for very large amounts of water. Thus a solution of a
given ammonia strength, it may be assumed, will always liberate the same
amount of heat when diluted with water, if the total amount of water after
dilution exceeds two hundred times the weight of ammonia present or there
abouts. This heat per pound of NHs contained will be designated as the
heat of complete dilution, and defined as the heat liberated when a solution
containing 1 Ib. of ammonia in a given amount of water is completely
diluted with water, or brought to the condition of 200 Ibs. or more of water per
pound of ammonia contained.
500
ENGINEERING THERMODYNAMICS
There is a rational relation between these three heats, that of complete
absorption, which is a constant, that of complete dilution, which depends only
on the original ratio of ammonia to water, and that of partial absorption,
which is a function of the character of solution receiving it, or if pure water
the amount. This relation is
J Heat of
complete
I absorption .
(Heat of partial absorption) =gg3
[ (+Heat of complete dilution) J
]b
Numerical values for heats of partial absorption are entirely lacking, but
Berthelot has given ten values for the heats of complete dilution for solutions
containing from 1 Ib. of water per pound of ammonia, to a little over one
hundred, but at only one temperature, 57 F. Up to Thomson's time these
figures seem to have been the sole dependence for engineering calculation;
he, however, added three more figures for more concentrated solutions, giving
not the heats of complete dilution, but those of partial dilution, that is, the
heat liberated when the water content is increased from one original value to
three different final values not corresponding to the state of complete dilu
tion. However, these may be used to check and correct the Berthelot figures
and are especially useful because they cover the doubtful range of his deter
mination and the range of ammonia strengths.
Berthelot's results are given in the following Table XXV and plotted in
Fig. 156, from which he derives a general law of relation given by the follow
ing formula Eq. (706), which is also plotted to show its agreement with the
points, and which is an equilateral hyperbola asymptotic to axes of H and w.
TABLE XXV
BERTHELOT'S DATA ON HEATS OF COMPLETE DILUTION OF AMMONIA
SOLUTIONS
Original Solu
tion, 1 Lb. Am
monia in (w)
Lbs. Water.
When Completely Diluted gives
Original Solu
tion, 1 Lb. Am
monia in (to)
Lbs. Water.
When Completely Diluted gives
1.04
136]
3.76
34]
1.06
134
6.11
22
1.13
1.98
124
51
B.T.U. per Ib. ammonia
10.06
57.39
2
B.T.U. per Ib. ammonia
3.18
41 J
116.47
OJ
142.5,
Heat of complete dilution, B.T.U. per Ib. NH 3 = (Berthelot) . . (706)
w
The agreement, it will be noted, is not very good for larger values of w than
4 or 5, which is unfortunate, as commercial ammonia lies between one part
ammonia to nine parts water, or w = 9, and one part ammonia to 39
HEAT AND MATTER
501
parts water, or w = 39. Nevertheless engineers have been using this formula
in these doubtful ranges for some years.
By means of the few but probably accurate figures given by Thomsen and
experimentally determined by him it is possible to check this practice. He
measured not the heat of complete dilution as did Berthelot, but the heats of
partial dilution, and the manner in which his data merge into those of Berthelot
make the combined results of doubly great value because of the difference in
method. Thomsen added a limited amount of water to a solution of ammonia
containing 3.39 Ibs. water per pound of ammonia and measured the heat,
which, of course, was not the heat of complete dilution, with the following
results:
THOMSEN'S DATA ON HEATS OF PARTIAL DILUTION OF AMMONIA
SOLUTIONS
Original Solution.
,' 1 Lb. Ammonia in (w) Lbs. Water.
Final Solution.
1 Lb. Ammonia in (w) Lbs. Water.
B.T.U. per Lb. Ammonia Absorbed
by the Ammonia Solution.
(3.39
10= ] 3.39
13.39
T19.27
w = \ 29.86
[56.33
+34
+37
+40
These results have been fitted into those of complete dilution by the relations
of Eqs. (707) and (708), and by the introduction of one assumption.
Heat of complete dilu
tion of original am
monia solution per
Ib. NH 3
Heat of partial dilution
from original to some
greater water content
per Ib. NH 3 .
Heat of complete
dilution of the
new solution per
Ib. NH 3 .
(707)
or
j [Heat of complete dilu1 ("Heat of complete dilu1
{ tion of new solution [ = \ tionof original solution \ 
[ per Ib. NH 3 J I per Ib. NH 3 .
Heat of partial dilution!
from original to final f (708)
solution per Ib. NH 3 . J
If the heat of complete dilution of the original solution containing 3.39 Ibs.
water per pound ammonia be taken from the Berthelot equation its value is
42, so assuming this to be correct it may be introduced in Eq. (708) as a constant,
.giving with the Thomsen figures the following:
Heat of complete dilution 
of new solution, per Ib. [
NH 3 .
{Heat of partial dilution j
from original to final r
solution per Ib. NH 8 . J
Heat of complete dilution per pound NH 3 with
(10 = 19.27) =4234 (Thomsen) = 8 ;
(w = 29.86) =4237 (Thomsen) = 5 ;
(to = 56.33) = 42  40 (Thomsen) = 2.
502
ENGINEERING THERMODYNAMICS
These three new points are also plotted and agree well with the Berthelot
equation, better even than the original Berthelot points themselves, so that
Thomsen's partial dilution figures seem to confirm Berthelot's complete dilution
data and the curve of Fig. 156 and Eq. (706) may be taken as truly represen
150
5 10 15 20 25 30
Pounds of Water Per Pound of Ammonia
1 K;. 156. Heat of Complete Dilution of Ammonia Water Solutions, by Excess Water.
tative of the heat of complete dilution of ammonia solutions and indirectly
of course, heats of partial dilutions as well.
Heats of absorption are more often needed in practical problems than heat
of complete or partial dilution, but these heats follow on the assumption that th'
heat of complete absorption, must be equal to the sum of the heat of partial absorp
tion, and the heat of complete dilution of the solution so formed.
HEAT AND MATTER
503
Hence
Heat of partial absorption
in w Ibs. water B.T.U.
per Ib. NH 3 .
Heat of complete
absorption in ex
cess water B.T.U.
per Ib. NH 3 .
[ Heat of complete dilution by
J excess water of solution
with w Ibs. water per Ib.
NH 3 , B.T.U. per Ib. NH 3 .
or
( Heat of partial absorption in water ^
I B.T.U. per Ib. NH 3 absorbed / "
142.5
w
(709)
800
a
I
400
200
Heat of par
ial absorption
112.5
W
Heat of complete
U8.5
V
Hea of complete abs<
dilutioi
rption
10
15 20 25 30
W =Lbs. Water per Ib. NH 3
FIG. 157. Ammoniawater Solutions, Relation between Heats
f Partial Absorption j
of \ Complete Absorption \ Shown Graphically.
I Complete Dilution /
It is interesting to note that the relation between these three quantities, heat
of partial absorption in limited amounts of water, heat of complete absorption
in excess water and heat of complete dilution of the solution in excess water,
can be shown graphically in Fig. 157, where the designations are self
explanatory.
Most important are the heats liberated when ammonia is absorbed not
by water but by weak ammonia solutions themselves, and these heats of partial
solution of ammonia in ammonia solutions are obtainable from the data already
established by a comparatively simple relation. In this case there are two
different solutions in question, an original one which becomes the second one
on receiving more ammonia. If the water received all the ammonia contained
in the second solution a certain quantity of heat would be liberated and it must
be equal to the total amount liberated by absorption of the first ammonia in
the water, and by the absorption of the second ammonia in the resulting solu
504
ENGINEERING THERMODYNAMICS
tion, whence this last quantity is obtainable by differences between the heats
of partial solution of ammonia in water alone.
Let w =lbs. water per Ib. ammonia in original solution which, therefore, consists of
w + 1 Ibs. solution, 1 Ib. of ammonia and w Ibs. of water.
A =lbs. ammonia added per Ib. ammonia already present, making new solution
containing w Ibs. of water and A+l Ibs. of ammonia or Ibs. water
Then
per Ib. ammonia in w+A +1 Ibs. of solution.
A+l
(Heat of partial absorption of 1 1^9 t
original 1 Ib. ammonia in w [ =893
Ibs. water, B.T.U.
fHeat of partial absorption of 1
\ all (A +1) Ibs. of ammonia I
in w Ibs. of water, B.T.U. J
w
Whence
Hjeat of partial
absorption of
A Ibs. NH 3 by
solution con
taining 1 Ib.
NH 3 in w Ibs.
water,B,T.U,
(710)
Therefore
Heat of partial absorption of A Ibs.
NH 3 by solution containing 1 Ib.
NH 3 in w Ibs. water B .T.U.per Ib.
NH 3
=893
(A +2)
(711)
As ammonia solution strengths are often given in terms of per cent of
ammonia present by weight and the heat of absorption in terms of changes in
the per cent strength, the following conversion factors are useful :
100
Per cent ammonia in original solution = Si = ,
IPHI
Per cent ammonia in final solution =82 = ,
Whence
100
Si
, (a)
ioo/ 2 i\
'' Si Viooftx
(712)
HEAT AND MATTER 505
These on substitution in Eq. (711) give the heat of absorption per pound
of ammonia absorbed to change the per cent NHs from Si to $2.
i Heat of partial absorption of 1 Ib. ) / <? \
of NH 3 in a solution containing V = 893 142.5 (^^rf ~^). . (713)
( lib. NH 3 in w Ibs. water, B.T.U. ) \ 1C ~ bl
When, however, the absorbed ammonia is to be driven off from the solution by
heating it, the discharge will consist partly of ammonia vapor and partly water
vapor, so the heat of liberation of a given amount of ammonia from solution will
be in practice that for the ammonia and equal to what would be liberated by its
absorption, but also in addition the heat of vaporization of the water vapor.
When, as in absorption refrigerating machine generators, the discharged
vapors meet incoming solution and are thereby partly condensed, prac
tically all except perhaps 2 per cent of the heat of vaporization of water vapor is
returned and the net heat of ammonia liberation is not materially different
from the value for absorption. If this is not done a correction must be intro
duced for the water vapor.
To assist in the solution of problems on the amount of noncondensible
ajases to be handled by dry vacuum pumps serving steam condensers, and on
she change in the composition of gases when scrubbed by water in the course
)f cooling and cleaning after manufacture, a table of solubilities of various gases
n water is added at the end of this Chapter in Tables LIII and LIV, as compiled
'pom various sources and reported in the Smithsonian physical tables. The
lumbers in the tables are volumes of standard gas, that is, gas measured at
2 F. and 1 atm. pressure, per unit volume of water, though they are at a
Afferent volume as absorbed or when absorbed at the temperature given,
jrhe pressure of the solution is in every case 29.92 ins. Hg., absolute and this
the combined pressure of both the gas and the water vapor.
Example. In the absorber of an ice machine of the absorption type, a weak solu
on of ammonia in water takes up the ammonia vapor coming from the refrigerating
ils, the heat found being removed by water. The weak liquor, as it is called, is con
nuously supplied and the rich liquor continually pumped away to the generator
here, by heating, some of the ammonia vapor is driven off to the condenser. As
iming the action in the absorber to be merely one of ammonia dissolving in a weak
>lution and that no water vapor leaves the generator, what will be the heat produced
the absorber and needed in the generator per pound of ammonia for the fol
wing assumed conditions:
Weak solution, 15 per cent NHs; strong solution, 30 per cent NHs;
temperature in absorber, 80 F.
From Eq. (713), the heat of absorption per Ib. of ammonia absorbed will be
506 ENGINEERING THERMODYNAMICS
where Si and 2 are the per cents of ammonia in weak and rich solutions, respec
tively, or
Q =893142.5^+^) =807. B.T.U. per Ib. NH 3 absorbed
\8o 70 /
Prob. 1. Ammonia is being absorbed by water at a temperature of 100 F. The
solution contains 10 per cent of ammonia. If the total pressure is 15 ins. of Hg, what
part of this is due to foreign gases, what part to ammonia and what part to water
vapor?
I^Prob. 2. How many cubic feet and how many pounds of the following gases can
be separately dissolved in 1000 gallons of pure water at atmospheric pressure and a
temperature of 50 F.? Air, carbon dioxide, and hydrogen. How would the results
be changed if the pressure were doubled? If the temperature were doubled?
Prob. 3. When either water or ammonia is added to an ammonia solution, heat
is evolved. Explain why this is so.
Prob. 4. Ammonia is being absorbed by a stream of running water, there being
5 Ibs. of water per pound of ammonia. What will be the heat developed per poun
of ammonia liquor formed?
Prob. 5. Ten pounds of the above solution receives an additional pound of am
monia. How much heat is generated by this action?
Prob. 6. A solution containing 10 per cent of ammonia receives an addition o
another 10 per cent. What was the amount of heat developed per pound of ammom
and per pound of solution when the second portion of the ammonia was added?
Prob. 7. The pressure in the absorber of an ammonia absorption machine is one a
mosphere and the temperature is 80 F. What is the maximum per cent of am
monia which can be absorbed by the water?
Prob. 8. The generator is working under a pressure of 125 Ibs. per square inc
gage and the heat is supplied by a steam coil in which the pressure is 30 Ibs. per squar
inch gage. What per cent of ammonia will be left in solution after passing throug
the generator and about how much steam must condense per ib. NHs discharged?
13. Combustion and Related Reactions. Relative Weights and Volume
of Substances and Elements, before and after Reaction. Not only ma
matter assume the three states of solid, liquid and vapor separately, in pair
simultaneously, or even all three together with various accompanying or causa
temperature, pressure, or heat content, conditions, without chemical chang
of the matter itself, as already discussed, but matter itself may change in kind
As Mellor puts it " matter appears to be endowed with properties in virtu
of which two or more dissimilar substances, when brought into close contact
give rise to other forms of matter possessing properties quite distinct from th
original substance." " The process of change is called a chemical reaction.
Chemical changes are assumed to be characterized by molecular rearrangemen
according to which molecules of elements may divide into atoms and th
separated atoms of one combine with those of another element, to form a molecul
of a new substance to be called a compound. Similarly, the molecules of con
pounds may split and reassociate, part of one, with part of another, to mak
a new compound or a single compound may split up into its elements whic
HEAT AND MATTER 507
remain separated, the last case being generally termed dissociation. All three
classes of change of substance are classifiable as chemical reactions, and there is
really no very rigid line to be drawn between the subclasses of reaction known
as combination and dissociation except when applied to the same substances,
in which case one is the reverse of the other. The complete or partial destruc
tion of a substance as such is commonly termed decomposition as, for example,
when hydrocarbon constituents of coal volatile, or liquid fuel, are changed by
excessively high temperature into free carbon, and other hydrocarbons or
even free hydrogen.
Every chemical change whether of combination or dissociation is accompanied
by a heat change of the system or group of substances. When the reaction
is such that heat is set free tending to raise the temperature of the reacting
mass unless it be carried away as liberated, the reaction is classed as exothermic.
On the contrary, those reactions that are accompanied by heat absorption,
tending to lower the temperature, unless heat be added to supply the absorp
tion, are classed as endothermic.
It appears then that every reaction tends to change the temperature of the
system, causing it to rise or fall according as the reaction is exothermic or
endothermic, except in the one case where several reactions occur simulta
neously, in which all the exothermic, set free just enough heat to supply what
is required for the endothermic.
The most important reaction in engineering is combustion, defined as the
chemical reaction between fuels and the oxygen of the air, which is exothermic
or heat liberating, and the source of practically all the heat used in engines for
conversion into work. Combustion is often classed as an oxidation process, and is
thus distinguished from another important engineering class of related reactions,
reduction or the reverse of oxidation, the most prominent case of which is
the change from carbon dioxide to carbon monoxide in gas producers, and in
a lesser degree in furnaces, according to which there is a reduction of oxygen
content per molecule, and which process is endothermic. The formation of
carbon monoxide directly from carbon by its oxidation, is sometimes defined
as partial combustion of carbon or incomplete oxidation because the product,
carbon monoxide, may further oxidise or burn to carbon dioxide by taking
up more oxygen. The substances produced by the partial combustion or partial
oxidation of one fuel element may also be considered as the result of the dis
sociation or reduction or deoxidation of the substances produced by its complete
combustion. Ordinarily, partial combustion and reduction are considered
as reverse processes producing the same substances by (a) exothermic reaction
of the primary substance in partial combustion and (6) the endothermic deoxi
dizing reaction of the products of complete exothermic reactions of the same
primary substances. It is also common to consider only the reaction of a sub
stance or socalled fuel element with oxygen, as combustion and other processes
whether involving oxygen or not, as related reactions. Thus, carbon combining
with oxygen to form carbonic acid is complete, and carbon combining with
k oxygen to form carbon monoxide incomplete combustion of carbon, while carbon
508 ENGINEERING THERMODYNAMICS
monoxide and steam reacting can hardly be considered as combustion, and is
best classified as a related reaction.
The number of elements entering into combustion and related reactions
is small, but the number of possible substances and reactions between them is
amazingly large, and the prediction of just what reactions will take place between
various groupings or mixtures of these substances extremely difficult and in some
cases quite impossible. The study of possible reactions has become the province
of physical chemistry, especially when the conditions controlling the result
are also subjects of study. These conditions include the temperature, pressure,
electrical state and the mutual relation of the elementary compounds present, and
the relation of these various conditions to the primary and resultant substances,
and the intermediate, successive, simultaneous or parallel reactions constitute the
subject matter of the study of chemical equilibrium. If chemical equilibrium
were better understood than it is, it would be possible to predict the resultant
from primary substances for specified conditions, but at the present time this
is impossible even though some of the greatest scientists the world has ever
known, have devoted their lives to the study and many volumes of specific
results have been published. Even if the exact prediction of the direction in
which reaction would proceed in an ordinary complex system and the extent
to which it would go in that direction, were made possible by a more complete
thermochemistry than now exists, it would not be of much use in engineering
because it is seldom possible to define the conditions that are present or to be met.
For example, in gas producers, solid coal, air and steam are the primary materials
and the product or result of their mutual reaction is a combustible gas. Engi
neers would like very much to be able to predict and control the exact com
position of this gas, but this is not possible nor will it in all probability ever be
possible, because it would first be necessary to fix the chemical and thermal
character of the coal, air and steam, to fix their relative quantities, to maintain
an absolutely uniform fuel bed as to size, porosity and, quality with some
other conditions equally elusive.
It must be understood, therefore, that while the possibility or even probability
of certain reactions taking place may be known, it is quite impossible to
predict just what will happen, or what products will result, when a given
group of primary substances mutually, react so that many important problems
of combustion in boiler furnaces and gas producers cannot be solved except by
approximation.
The approximation takes the form of a calculation which is exact, based on
an hypothesis which does not represent the facts of the case. In other words,
engineering calculations about combustion are always to be prefaced by a
statement that certain substances are going to change completely or within
a given degree to certain others, whether they will or not. Furthermore, the
substances must be defined chemically by their symbols, and then will it be
possible to calculate the relative weights and volumes of the various substances
that can so react, the corresponding relative weights and volumes of the products,
and the heat liberated or absorbed, but not otherwise. It is quite important
HEAT AND MATTER 509
that too much confidence be not put in these results, which are no more correct
i than is the assumption of what substances are to be formed.
From the study of chemical equilibrium a few principles of guidance have
ibeen developed that help a little but not very much. For example, Van't
! Hoff's " law of movable equilibrium " says that when the temperature of a system
 is raised that reaction takes place which is accompanied by absorption of heat
[and conversely. Another similar law is to the effect that a rise of pressure in
jja system in equilibrium causes that reaction that is accompanied by a reduction
jof volume. There are more of this sort but they are entirely too general to make
[at possible to avoid the procedure adopted by engineers of assuming the kind
bf reaction, and then calculating quantities and heats that should and would
iccompany it, if it did take place and, however, crude this may look it is very
useful, and in most cases good enough or rather as good as the knowledge of other
Conditions to be met.
Carbon and hydrogen are the only chemical elements of fuel value, and all
Commercial fuels, including wood, peat, lignite, bituminous and anthracite
ifcoals, charcoal, coke, crude petroleum oils with their distillates, gasolene,
ikerosene, and their residues, tars, heavy oils, alcohols, benzole a bituminous
tji'.oal product, natural gas as well as blast furnace, carburetted and uncar
'Jmretted water gas, coal gas, producer gas, and oil gas, are compounds,
Mixtures and mixtures of compounds of these fuel elements, with oxygen
'.la some cases. The one exception is sulphur, which exists in many solid,
aquid and gaseous fuels as an impurity, but which also has some small
. uel value.
. i This being the case, the number of products to be formed by the complete
jombustion of fuel is also small, and includes only carbon dioxide and water
i ! apor, with the nitrogen carried by the air, and a small amount of sulphur
iompounds, often ignored.
The process of reaction, whether combustion or one of the related ones,
I* to be described by the usual chemical equation which has the additional
gnificance of showing the relative weights involved directly, because,
(a) The total number of atoms of each chemical element on each side of
the equation must be the same.
(b) The sum of the products of the atomic weight of each atom and the
number present, must be the same on each side of the equation.
This is the same as saying, (a) that the total weight of hydrogen in the prod
3ts must be the same as the total weight of hydrogen in the original mixture
id so also for the other elements, and (b) the total weight of the original mixture
' reacting substances must be the same as that of the products.
Natural fuels while sometimes consisting of simple elements like carbon
hydrogen alone, or simple compounds like carbon monoxide or methane alone,
'e more often mixtures. Their reaction equations are then to be derived
om combinations of the equations representing the reactions of elements and
compounds with oxygen or with each other. To facilitate this the following
ble of some characteristic reactions of simple substances is inserted :
510
ENGINEERING THERMODYNAMICS
SOME COMBUSTION AND RELATED REACTIONS OF FUEL ELEMENTS AND
COMPOUNDS
Combustion o
f f uelel ements J c
Product of
partial car
bon combus
tion
!H 2 fO 2 2H 2 O .....
V A /
(2)
(3)
2COO 2 2CO 2 . . .
Paraffin series CnHjw+2:
Methane CH 4 +2O 2 = CO 2 +2H 2 O . . .
(4)
Combustion
of
Chemical
Compounds
Hydrocarbons
Ethane C S H 6 +O 2
J* 1 X~ TT 1 /
= 2CO 2 +3H 2 O . . (5)
) O 2 = wCO 2 + (n + 1 )H 2 O (6)
O 2 =2CO 2 +2H 2 O . . (7)
2 =3CO 2 +3H 2 O . . (8)
General C w H 2 n+ 2 + (^
Olefines or Ethylene series C re H 2w :
Ethylene CzK
Propylene C 3 H 6 +
General C K H 2n + ^<X = nCO 2 +nH 2 O . .
(9)
Alcohols
Monatomic alcohols CnH2n+iOH:
Methyl alcohol CH 3 OH+o 2 = CO 2 +2H 2 O . .
Ethyl alcohol C 2 H 5 OH+3O 2 =2CO 2 +3H 2 O. .
General CwH^ +iOH + O 2 = nCO 2 + (n + 1 )H 2 O
(10)
(11)
(12)
Related Reactions, Incomplete Combustion, Decom
position, Dissociation.
2C+0 2
CO 2 +H 2 =
CH4+CO+H 2
C+CO 2 =
CO+H 2 O =
2CO 2 =
2H 2 O =
CH 4
C 2 H 4 =
= 2CO .... (13;
= CO+H 2 O . . (14
= C 2 H 4 +H 2 O. . (15
= 2CO (16:
:C0 2 +H 2 .
= 2CO+O 2 .
= 2H 2 +O 2 .
= C+2H 2 .
= C+CH 4 .
2CH 4
3C 2 H 2 =
C 2 H 2 +3H 2
C 2 H 4 +2C 2 H 2
(17
.(18
.(19
.(20
.(21
.(22,
.(23
. (24
All these and any other reactions are characterized by definite weigh
relations which are given directly by the reaction equation by introducing th
weight of each element from a table of atomic weights, and for this purpos
the nearest whole number is close enough. For example, the complete com
bustion of hexane, CeHn, the main constituent of gasolene, to carbon dioxid
and water, is given by,
or
C 6 H4+f02 = 6CO2+7H 2
HEAT AND MATTER 5H
This may be interpreted as follows, taking the atomic weight of C = 12 of
H=l, and of O = 16.
(6X12 = 72) Ibs. carbon
+ (14 XI =.14) Ibs. hydrogen
+ (9^X2X16 = 304) Ibs. oxygen
make
(6X12 = 72) Ibs. carbon
+ (6X2X16 = 192) Ibs. oxygen
+ (7X2X1 = 14) Ibs. hydrogen
+ (7X16 = 112) Ibs. oxygen
or
, iLt^u =86) lbs ' i hexane \ make / (72+192 = 264 Ibs. carbon dioxide
+ (304) Ibs. oxygen e \ +(14+112 = 126) Ibs. water vapor
This general weight relation may be simplified somewhat by considering
the weight of each one of the substances as unity, thus yielding the following
different, but equivalent interpretations :
^1 Ib. hexane+ Ibs. oxygen j make ( Ibs. car. diox.+i^ Ibs. water.)
f 1 Ib. oxy. + Ibs. hexane J make ( Ibs. car. diox.+^ Ibs. water.)
/ 12fi x result from com,
1 Ib. carb. diox.+ Ibs.water ) plete combus(^. i bs> hex.+^ Ibs. oxy. )
^o4 / tion of \*o4 264 * /
2fi4 x result from com,
Ib. water+^ Ibs. car. diox. ) plete combus/^ [b s . hex.+^ Ibs. oxy. )
i^b / tion of . V 1 ^" l^o /
It might also be said from the same data that,
1 Ib. hexane completely burned with oxygen yields 390 Ibs. products.
1 Ib. oxygen will completely burn ^ Ibs. hexane and yield 390 Ibs. products.
This particular example can be analyzed in still another way, yielding a general
expression for the combustion of an analyzed fuel.
Fuel analyses are reported in two ways: (a) Proximate, giving per cent of
each independent compound, or separately determined constituent substance;
(b) ultimate, giving the per cent of each chemical element. Applying this dis
tinction to the original mixture of hexane and oxygen, and its products the
proximate analyses are:
Original mixture
Hexane = 390
304
Oxygen =^ = 77.9 % by weight.
Total 100.00%
512
ENGINEERING THERMODYNAMICS
Products
264
Carbon dioxide =5Hn
67.6 % by weight.
126
Water vapor or liquid = ^ = 23.4 % by weight.
Total 100.00%
Using the ultimate analysis method of designation for mixtures and products
72
Original mixture
Products
Carbon (in hexane)
Hydro, (in hexane)
Oxygen (free)
Total
Carbon (in C0 2 )
Hydrogen (water)
Oxygen (in C0 2 )
Oxygen (water)
Total
14
390
304
: 390
72
14
390
192
: 390
112
: 390
= 18.4% by weight.
= 3.6 % by weight.
= 78.0 % by weight.
100.00%
= 18.4 % by weight.
= 3.6 % by weight.
fjj 78.0 % by weight.
100.00%
Neither combustion or its related reactions take place with oxygen alone,
but with air containing oxygen, 23.2 per cent, and nitrogen, 76.8 per cent
76 8
by weight, each pound of oxygen carrying with it ^ = 3.31 Ibs. of nitrogen
2i*j.2i
or existing in 4.31 Ibs. of air. The nitrogen is generally considered neutral,
though it may form compounds with hydrogen, such as ammonia directly,
or with oxygen, such as nitrous oxide if conditions are right. If neutral, it
has the effect of changing the weight of the mixture by an amount depending
on the proportion of oxygen that came from air.
/ (Weight of mixture with oxygen alone) \ / i . r .  1 \
' +3.31 (wt. oxygen present) } = < wei e ht of mlxture wth alr) 
or
air is used instead of oxygen)
 3.31 (wt. of oxygen present).
HEAT x\ND MATTER
513
This will add in the present case 304X3.31 = 1006 Ibs. of nitrogen in the
combustion of one molecule or 86 Ibs. of hexane, the sum of 304 Ibs. of oxygen
and 1006 Ibs. of nitrogen, giving 1310 Ibs., the weight of air required, and rais
ing the total weight of the mixture to 1396 Ibs. Of course, the per cent of the
various constituents of the mixture and products is now changed, but the
amount of change is quickly computed. All these relative numbers can be
conveniently given in tables of conversion factors, such a table for hexane
being given below, as a type form, useful in e veryday work when it is nec
essary to make repeated calculations with some one fuel.
CONVERSION FACTORS FOR WEIGHTS IN THE COMPLETE COMBUSTION OF
HEXANE WITH AIR TO WATER AND CARBON DIOXIDE
Original
! Mixture.
Final
Mixture.
Hexane
C 6 Hi4.
Oxygen
Nitrogen
N.
Air.
Carbon Di
oxide CO 2 .
Water
HzO.
1396.
1396.
86.
304.
1006.
1310.
264.
126.
1.
1.
.0616
.218
.720 .938
.189
.091
16.22
16.22
1.
3.54
11.68 15.22
3.07
1.46
4.59
4.59
.283
1.
3.31 4.31
.87
.41
1.387
1.387
.085
.302
1.
1.302
.262
.125
1.066
1.066
.066
.232
.768
1.
.202
.096
5.29
5.29
.326
1.15
3.81
4.96
1.
.48
11.08
11.08
.683
2.42
7.98
10.40
2.10
1.
The average analysis of pure air is given by Parkes as being 02 = 20.96 per cent;
J0 2 = .04 per cent; N2 = 76.00 per cent by volume, giving 02 = 23.20 per cent;
302 = .06 per cent; N2 = 76.74 per cent by weight. Regarding the C02 as
Deing negligible, the relation may be used for the purpose of computations of
ihis sort, O2 = 23.2 per cent and N2 = 76.8 per cent. On the assumption that
my nitrogen present came from air and not from any other compound, such
is ammonia, it must have been associated with oxygen in the proportion 3.31
X) 1 of O2. The weight of oxygen on both sides must be equal and the weight
)f air on one side must be 4.31 times the weight of nitrogen. Of the nitrogen
,hat is present X(the weight of nitrogen present) must have come from air
o.ol
tnd the rest from the other substances.
A muchused weight relation is, for the weight of air and products per pound
)f fuel, and on the assumption that
H = part of hydrogen by weight per pound fuel,
C = part of carbon by weight per pound fuel,
00
Weight of air per Ib. of fuel)=CXYx4,31+HX8X4.31 = 11.49C+34.46H,(7i:)
514 ENGINEERING THERMODYNAMICS
32
Weight of products per Ib. fuel = 3C Ibs. C0 2 +9H Ibs. H 2 O+^
X3.31C Ibs. N+8X3.31H Ibs. N
= 3.67C in form of CO 2 +9H in form of H 2 O+8.82 C in form of N
= 12.49C+35.46H, ................ (715
for complete combustion in air with no more air supplied than enters into th<
reaction
Most of the practical problems concerning the relative amounts of substance;
involved in combustion and reduction processes are concerned with gases, a
least on one side of the equation and sometimes on both. Direct combustioi
and the gasification of fuels in producers and complete combustion in boilei
furnaces always yields gas mixtures, the composition of which is always volui
metrically determined by analysis, while the explosive mixtures or priman
working substances of gas engines are gaseous initially and remain so afte:j
combustion. It is quite necessary, therefore, to transform the weight relation;
of the reaction equation into another form yielding volumes. There are thre
ways of doing this, all equivalent and all yielding the same result if the constant'
used are consistent, and if the gases and vapors follow the Avagadro law, oi
which the most useful of the methods depends.
1. The volume at standard conditions of any substance can be found fron
the weight present by multiplying that weight by the specific volume of th'
substances ; in cubic feet per pound, at standard condition of 1 atm. at 32 F.
2. The molecular weight in pounds of any gaseous or vapor substanc
assumed to follow Avagadro's law occupies 358 cu.ft.
3. The volumetric relations are given directly by the coefficients of the sub
stance in the chemical equation when the .substances are gaseous and all ente
into the reaction, and when each is expressed in terms of molecules present.
The first method needs no explanation or further development and the secom
and third are really one. // in the reaction equations there be written m
any substance its molecular weightX358, the product mil be the volume c<
that substance in the reaction in cubic fe